CBSE NCERT Solutions for Class 12 Maths Chapter 04€¦ · Class-XII-Maths Determinants 1 Practice...



Now, |𝑥2 − 𝑥 + 1 𝑥 − 1𝑥 + 1 𝑥 + 1

|

= (𝑥2 − 𝑥 + 1) × (𝑥 + 1) − (𝑥 − 1) × (𝑥 + 1)

= 𝑥3 + 𝑥2 − 𝑥2 − 𝑥 + 𝑥 + 1 − (𝑥2 − 1)

= 𝑥3 − 𝑥2 + 2

Hence, |𝑥2 − 𝑥 + 1 𝑥 − 1𝑥 + 1 𝑥 + 1

| = 𝑥3 − 𝑥2 + 2

3. If 𝐴 = [1 24 2

], then show that |2𝐴| = 4|𝐴|

Solution:

Given that 𝐴 = [1 24 2

]

Now, 2𝐴 = [2 × 1 2 × 22 × 4 2 × 2

]

= [2 48 4

]

|2𝐴| = |2 48 4

|


= 2 × 4 − 4 × 8 = 8 − 32 = −24 …(i)

And 4|𝐴| = 4 |1 24 2

|


= 4(1 × 2 − 2 × 4) = 4(−6) = −24 …(ii)

From the equation (i) and (ii),

we get |2𝐴| = 4|𝐴|

Hence proved.

4. If 𝐴 = [1 0 10 1 20 0 4

], then show that |3𝐴| = 27|𝐴|



Solution:

Given that 𝐴 = [1 0 10 1 20 0 4

]

⇒ 3𝐴 = [3 0 30 3 60 0 12

]

Now, |3𝐴| = |3 0 30 3 60 0 12

|


= 3(36 − 0) − 0(0 − 0) + 3(0 − 0) = 108 …(i)

And 27|𝐴| = 27 |1 0 10 1 20 0 4

|


= 27{1(4 − 0) − 0(0 − 0) + 1(0 − 0)} = 27(4) = 108 …(ii)

From the equation (i) and (ii),

we get |3𝐴| = 27|𝐴|

Hence proved.

5. Evaluate the determinants:

(i) |3 −1 −20 0 −13 −5 0

|

(ii) |3 −4 51 1 −22 3 1

|

(iii) |0 1 2

−1 0 −3−2 3 0

|

(iv) |2 −1 −20 2 −13 −5 0

|

Solution:

(i) Given determinant is |3 −1 −20 0 −13 −5 0

|



Now, expanding the determinant along 𝑅1, we get

|3 −1 −20 0 −13 −5 0

| = 3(0 − 5) + 1(0 + 3) − 2(0 − 0) = −15 + 3 − 0 = −12

Hence |3 −1 −20 0 −13 −5 0

| = −12

Given determinant is |3 −4 51 1 −22 3 1

|


|3 −4 51 1 −22 3 1

| = 3(1 + 6) + 4(1 + 4) + 5(3 − 2) = 21 + 20 + 5 = 46

Hence |3 −4 51 1 −22 3 1

| = 46

(iii) Given determinant is |0 1 2

−1 0 −3−2 3 0

|


|0 1 2

−1 0 −3−2 3 0

| = 0(0 + 9) − 1(0 − 6) + 2(−3 − 0) = 0 + 6 − 6

So, the given determinant |0 1 2

−1 0 −3−2 3 0

| = 0

Given determinant is |2 −1 −20 2 −13 −5 0

|


|2 −1 −20 2 −13 −5 0

| = 2(0 − 5) + 1(0 + 3) − 2(0 − 6) = −10 + 3 + 12 = 5

Hence, |2 −1 −20 2 −13 −5 0

| = 5

6. If 𝐴 = [1 1 −22 1 −35 4 −9

], find |𝐴|.



Solution:

Given that 𝐴 = [1 1 −22 1 −35 4 −9

]

Now, |𝐴| = |1 1 −22 1 −35 4 −9

|

Expanding the determinant along 𝑅1, we get

= 1(−9 + 12) − 1(−18 + 15) − 2(8 − 5) = 3 + 3 − 6 = 0

Hence, |𝐴| = 0

7. Find values of 𝑥, if

(i) |2 45 1

| = |2𝑥 46 𝑥

|

(ii) |2 34 5

| = |𝑥 32𝑥 5

|

Solution:

(i) Given that |2 45 1

| = |2𝑥 46 𝑥

|

⇒ 2 − 20 = 2𝑥2 − 24

⇒ 𝑥2 = 3

⇒ 𝑥 = ±√3

Hence, the values of 𝑥 are ± √3

(ii) Given that |2 34 5

| = |𝑥 32𝑥 5

|

⇒ 10 − 12 = 5𝑥 − 6𝑥

⇒ −2 = −𝑥

⇒ 𝑥 = 2

Hence, the value of 𝑥 is 2

8. If |𝑥 218 𝑥

| = |6 218 6

|, then 𝑥 is equal to:

(A) 6



6. |0 𝑎 −𝑏

−𝑎 0 −𝑐𝑏 𝑐 0

| = 0

Solution:

𝐿𝐻𝑆 = |0 𝑎 −𝑏

−𝑎 0 −𝑐𝑏 𝑐 0

|

= |0 𝑎 −𝑏

−𝑎𝑏 0 −𝑏𝑐𝑎𝑏 𝑎𝑐 0

| [Applying 𝑅2 → 𝑏𝑅2 and R3 → 𝑎𝑅3]

= |0 𝑎 −𝑏0 𝑎𝑐 −𝑏𝑐𝑎𝑏 𝑎𝑐 0

| [Applying 𝑅2 → 𝑅2 + 𝑅3]

= 𝑎𝑏(−𝑎𝑏𝑐 + 𝑎𝑏𝑐) [Expanding along 𝐶1]

= 𝑎𝑏(0) = 0 = 𝑅𝐻𝑆

Hence, LHS = RHS

7. |−𝑎2 𝑎𝑏 𝑎𝑐𝑏𝑎 −𝑏2 𝑏𝑐𝑐𝑎 𝑐𝑏 −𝑐2

| = 4𝑎2𝑏2𝑐2

Solution:

𝐿𝐻𝑆 = |−𝑎2 𝑎𝑏 𝑎𝑐𝑏𝑎 −𝑏2 𝑏𝑐𝑐𝑎 𝑐𝑏 −𝑐2

|

= 𝑎𝑏𝑐 |−𝑎 𝑎 𝑎𝑏 −𝑏 𝑏𝑐 𝑐 −𝑐

| [Taking a, b, c as common from C1, 𝐶2, 𝐶3 respectively]

= 𝑎2𝑏2𝑐2 |−1 1 11 −1 11 1 −1

| [Taking a, b, c as common from R1, 𝑅2, 𝑅3 respectively]

= 𝑎2𝑏2𝑐2 |0 1 10 −1 12 1 −1

| [Applying 𝐶1 → 𝐶1 + 𝐶2]

= 𝑎2𝑏2𝑐2{2(1 + 1)} [Expanding along 𝐶1]

= 4𝑎2𝑏2𝑐2 = 𝑅𝐻𝑆

Hence, LHS = RHS



8. Using properties of determinants, show that

(i) |1 𝑎 𝑎2

1 𝑏 𝑏2

1 𝑐 𝑐2

| = (𝑎 − 𝑏)(𝑏 − 𝑐)(𝑐 − 𝑎)

(ii) |1 1 1𝑎 𝑏 𝑐𝑎3 𝑏3 𝑐3

| = (𝑎 − 𝑏)(𝑏 − 𝑐)(𝑐 − 𝑎)(𝑎 + 𝑏 + 𝑐)

Solution:

(i) 𝐿𝐻𝑆 = |1 𝑎 𝑎2

1 𝑏 𝑏2

1 𝑐 𝑐2

|

= |0 𝑎 − 𝑏 𝑎2 − 𝑏2

0 𝑏 − 𝑐 𝑏2 − 𝑐2

1 𝑐 𝑐2

| [Applying R1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]

= (𝑎 − 𝑏)(𝑏 − 𝑐) |0 1 𝑎 + 𝑏0 1 𝑏 + 𝑐1 𝑐 𝑐2

| [Taking common a − b from R1and b − c from R2] ]

= (𝑎 − 𝑏)(𝑏 − 𝑐){1(𝑏 + 𝑐 − 𝑎 − 𝑏)} [Expanding along 𝐶1]

= (𝑎 − 𝑏)(𝑏 − 𝑐)(𝑐 − 𝑎) = 𝑅𝐻𝑆

Hence, LHS = RHS

(ii) |1 1 1𝑎 𝑏 𝑐𝑎3 𝑏3 𝑐3

| = (𝑎 − 𝑏)(𝑏 − 𝑐)(𝑐 − 𝑎)(𝑎 + 𝑏 + 𝑐)

𝐿𝐻𝑆 = |1 1 1𝑎 𝑏 𝑐𝑎3 𝑏3 𝑐3

|

= |0 0 1

𝑎 − 𝑏 𝑏 − 𝑐 𝑐𝑎3 − 𝑏3 𝑏3 − 𝑐3 𝑐3

| [By C1 → 𝐶1 − 𝐶2, 𝐶2 → 𝐶2 − 𝐶3]

= (𝑎 − 𝑏)(𝑏 − 𝑐) |0 0 11 1 𝑐

𝑎2 + 𝑎𝑏 + 𝑏2 𝑏2 + 𝑏𝑐 + 𝑐2 𝑐3| [Taking common a − b from C1and b − c from C2]

= (𝑎 − 𝑏)(𝑏 − 𝑐){1(𝑏2 + 𝑏𝑐 + 𝑐2) − (𝑎2 + 𝑎𝑏 + 𝑏2)} [Expanding along 𝑅1]

= (𝑎 − 𝑏)(𝑏 − 𝑐){𝑐2 − 𝑎2 + 𝑏𝑐 − 𝑎𝑏}

= (𝑎 − 𝑏)(𝑏 − 𝑐){(𝑐 − 𝑎)(𝑐 + 𝑎) + 𝑏(𝑐 − 𝑎)}

= (𝑎 − 𝑏)(𝑏 − 𝑐)(𝑐 − 𝑎){𝑐 + 𝑎 + 𝑏}

= (𝑎 − 𝑏)(𝑏 − 𝑐)(𝑐 − 𝑎)(𝑎 + 𝑏 + 𝑐) = 𝑅𝐻𝑆

Hence, LHS = RHS




|

𝑥 𝑥2 𝑦𝑧

𝑦 𝑦2 𝑧𝑥

𝑧 𝑧2 𝑥𝑦

| = (𝑥 − 𝑦)(𝑦 − 𝑧)(𝑧 − 𝑥)(𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥)

Solution:

𝐿𝐻𝑆 = |

𝑥 𝑥2 𝑦𝑧

𝑦 𝑦2 𝑧𝑥

𝑧 𝑧2 𝑥𝑦

|

= |

𝑥2 𝑥3 𝑥𝑦𝑧

𝑦2 𝑦3 𝑥𝑦𝑧

𝑧2 𝑧3 𝑥𝑦𝑧

| [Applying 𝑅1 → 𝑥𝑅1, 𝑅2 → 𝑦𝑅2, 𝑅3 → 𝑧𝑅3]

= 𝑥𝑦𝑧 |𝑥2 𝑥3 1𝑦2 𝑦3 1

𝑧2 𝑧3 1

| [Taking xyz as common from C3]

= 𝑥𝑦𝑧 |𝑥2 − 𝑦2 𝑥3 − 𝑦3 0

𝑦2 − 𝑧2 𝑦3 − 𝑧3 0

𝑧2 𝑧3 1

| [Applying R1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]

= 𝑥𝑦𝑧(𝑥 − 𝑦)(𝑦 − 𝑧) |𝑥 + 𝑦 𝑥2 + 𝑥𝑦 + 𝑦2 0

𝑦 + 𝑧 𝑦2 + 𝑦𝑧 + 𝑧2 0

𝑧2 𝑧3 1

| [Taking 𝑥 − 𝑦 as common from R1and y − z from R2]

= 𝑥𝑦𝑧(𝑥 − 𝑦)(𝑦 − 𝑧){(𝑥 + 𝑦)(𝑦2 + 𝑦𝑧 + 𝑧2) − (𝑦 + 𝑧)(𝑥2 + 𝑥𝑦 + 𝑦2)} [Expanding along C3]

= 𝑥𝑦𝑧(𝑥 − 𝑦)(𝑦 − 𝑧){𝑥𝑦2 + 𝑥𝑦𝑧 + 𝑥𝑧2 + 𝑦3 + 𝑦2𝑧 + 𝑦𝑧2 − (𝑥2𝑦 + 𝑥𝑦2 + 𝑦3 + 𝑥2𝑧 +

𝑥𝑦𝑧 + 𝑦2𝑧)}

= 𝑥𝑦𝑧(𝑥 − 𝑦)(𝑦 − 𝑧){𝑥𝑧2 + 𝑦𝑧2 − 𝑥2𝑦 − 𝑥2𝑧}

= 𝑥𝑦𝑧(𝑥 − 𝑦)(𝑦 − 𝑧){𝑥𝑧2 − 𝑥2𝑧 + 𝑦𝑧2 − 𝑥2𝑦}

= 𝑥𝑦𝑧(𝑥 − 𝑦)(𝑦 − 𝑧){𝑥𝑧(𝑧 − 𝑥) + 𝑦(𝑧2 − 𝑥2)}

= 𝑥𝑦𝑧(𝑥 − 𝑦)(𝑦 − 𝑧)(𝑧 − 𝑥){𝑥𝑧 + 𝑦(𝑧 + 𝑥)}

= (𝑥 − 𝑦)(𝑦 − 𝑧)(𝑧 − 𝑥)(𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥) = 𝑅𝐻𝑆

Hence, LHS = RHS




(i) |𝑥 + 4 2𝑥 2𝑥2𝑥 𝑥 + 4 2𝑥2𝑥 2𝑥 𝑥 + 4

| = (5𝑥 + 4)(4 − 𝑥)2

(ii) |

𝑦 + 𝑘 𝑦 𝑦𝑦 𝑦 + 𝑘 𝑦𝑦 𝑦 𝑦 + 𝑘

| = 𝑘2(3𝑦 + 𝑘)

Solution:

(i) |𝑥 + 4 2𝑥 2𝑥2𝑥 𝑥 + 4 2𝑥2𝑥 2𝑥 𝑥 + 4

| = (5𝑥 + 4)(4 − 𝑥)2

𝐿𝐻𝑆 = |𝑥 + 4 2𝑥 2𝑥2𝑥 𝑥 + 4 2𝑥2𝑥 2𝑥 𝑥 + 4

|

= |5𝑥 + 4 2𝑥 2𝑥5𝑥 + 4 𝑥 + 4 2𝑥5𝑥 + 4 2𝑥 𝑥 + 4


= (5𝑥 + 4) |1 2𝑥 2𝑥1 𝑥 + 4 2𝑥1 2𝑥 𝑥 + 4

| [Taking 5x + 4 as common from C1]

= (5𝑥 + 4) |0 𝑥 − 4 00 4 − 𝑥 𝑥 − 41 2𝑥 𝑥 + 4

| [Applying R1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]

= (5𝑥 + 4){(𝑥 − 4)(𝑥 − 4) − (4 − 𝑥)0} [Expanding along C1]

= (5𝑥 + 4)(4 − 𝑥)2 = 𝑅𝐻𝑆

Hence, LHS = RHS

(iii) |


| = 𝑘2(3𝑦 + 𝑘)

𝐿𝐻𝑆 = |


|

= |

3𝑦 + 𝑘 𝑦 𝑦3𝑦 + 𝑘 𝑦 + 𝑘 𝑦3𝑦 + 𝑘 𝑦 𝑦 + 𝑘

| [Applying C1 → 𝐶1 + 𝐶2 + 𝐶3]

= (3𝑦 + 𝑘) |

1 𝑦 𝑦1 𝑦 + 𝑘 𝑦1 𝑦 𝑦 + 𝑘

| [Taking 3y + k as common from C1]

= (3𝑦 + 𝑘) |0 −𝑘 00 𝑘 −𝑘1 𝑦 𝑦 + 𝑘

| [Applying R1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]



= (3𝑦 + 𝑘){(−𝑘)(−𝑘) − (𝑘)0} [Expanding along C1]

= (3𝑦 + 𝑘)𝑘2 = 𝑅𝐻𝑆

Hence, LHS = RHS


(i) |𝑎 − 𝑏 − 𝑐 2𝑎 2𝑎

2𝑏 𝑏 − 𝑐 − 𝑎 2𝑏2𝑐 2𝑐 𝑐 − 𝑎 − 𝑏

| = (𝑎 + 𝑏 + 𝑐)3

(ii) |

𝑥 + 𝑦 + 2𝑧 𝑥 𝑦𝑧 𝑦 + 𝑧 + 2𝑥 𝑦𝑧 𝑥 𝑧 + 𝑥 + 2𝑦

| = 2(𝑥 + 𝑦 + 𝑧)3

Solution:

(i) 𝐿𝐻𝑆 = |𝑎 − 𝑏 − 𝑐 2𝑎 2𝑎

2𝑏 𝑏 − 𝑐 − 𝑎 2𝑏2𝑐 2𝑐 𝑐 − 𝑎 − 𝑏

|

= |𝑎 + 𝑏 + 𝑐 𝑎 + 𝑏 + 𝑐 𝑎 + 𝑏 + 𝑐

2𝑏 𝑏 − 𝑐 − 𝑎 2𝑏2𝑐 2𝑐 𝑐 − 𝑎 − 𝑏

| [Applying R1 → 𝑅1 + 𝑅2 + 𝑅3]

= (𝑎 + 𝑏 + 𝑐) |1 1 12𝑏 𝑏 − 𝑐 − 𝑎 2𝑏2𝑐 2𝑐 𝑐 − 𝑎 − 𝑏

| [Taking a + b + c as commong from R1]

= (𝑎 + 𝑏 + 𝑐) |0 0 1

𝑎 + 𝑏 + 𝑐 −𝑎 − 𝑏 − 𝑐 2𝑏0 𝑎 + 𝑏 + 𝑐 𝑐 − 𝑎 − 𝑏

| [By C1 → 𝐶1 − 𝐶2, 𝐶2 → 𝐶2 − 𝐶3]

= (𝑎 + 𝑏 + 𝑐){(𝑎 + 𝑏 + 𝑐)2 − 0} [Expanding along R1]

= (𝑎 + 𝑏 + 𝑐)3 = 𝑅𝐻𝑆

(𝑖𝑖) |


| = 2(𝑥 + 𝑦 + 𝑧)3

𝐿𝐻𝑆 = |


|

= |

2(𝑥 + 𝑦 + 𝑧) 𝑥 𝑦

2(𝑥 + 𝑦 + 𝑧) 𝑦 + 𝑧 + 2𝑥 𝑦

2(𝑥 + 𝑦 + 𝑧) 𝑥 𝑧 + 𝑥 + 2𝑦

| [Applying C1 → C1 + C2 + C3]



= 2(𝑥 + 𝑦 + 𝑧) |

1 𝑥 𝑦1 𝑦 + 𝑧 + 2𝑥 𝑦1 𝑥 𝑧 + 𝑥 + 2𝑦

| [Taking 2(𝑥 + 𝑦 + 𝑧) common from C1]

= 2(𝑥 + 𝑦 + 𝑧) |

0 −(𝑥 + 𝑦 + 𝑧) 0

0 𝑥 + 𝑦 + 𝑧 −(𝑥 + 𝑦 + 𝑧)1 𝑥 𝑧 + 𝑥 + 2𝑦

| [By R1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]

= 2(𝑥 + 𝑦 + 𝑧){(𝑥 + 𝑦 + 𝑧)2 − 0} [Expanding along C1]

= 2(𝑥 + 𝑦 + 𝑧)3 = 𝑅𝐻𝑆

Hence, LHS = RHS


|1 𝑥 𝑥2

𝑥2 1 𝑥𝑥 𝑥2 1

| = (1 − 𝑥3)2

Solution:

𝐿𝐻𝑆 = |1 𝑥 𝑥2

𝑥2 1 𝑥𝑥 𝑥2 1

|

= |1 + 𝑥 + 𝑥2 𝑥 𝑥2

1 + 𝑥 + 𝑥2 1 𝑥1 + 𝑥 + 𝑥2 𝑥2 1

| [Applying C1 → 𝐶1 + 𝐶2 + 𝐶3]

= (1 + 𝑥 + 𝑥2) |1 𝑥 𝑥2

1 1 𝑥1 𝑥2 1

| [Taking 1 + 𝑥 + 𝑥2 as common from C1]

= (1 + 𝑥 + 𝑥2) |0 𝑥 − 1 𝑥2 − 𝑥0 1 − 𝑥2 𝑥 − 11 𝑥2 1

| [Applying R1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]

= (1 + 𝑥 + 𝑥2)(1 − 𝑥)2 |0 −1 −𝑥0 1 − 𝑥 −11 𝑥2 1

| [Taking 1 − 𝑥 as common from R1 and R2]

= (1 + 𝑥 + 𝑥2)(1 − 𝑥)2{1 + 𝑥(1 + 𝑥)} [Expanding along C1]

= (1 + 𝑥 + 𝑥2)(1 − 𝑥2)(1 + 𝑥 + 𝑥2) = (1 − 𝑥3)2 = 𝑅𝐻𝑆

Hence, LHS = RHS




|1 + 𝑎2 − 𝑏2 2𝑎𝑏 −2𝑏

2𝑎𝑏 1 − 𝑎2 + 𝑏2 2𝑎2𝑏 −2𝑎 1 − 𝑎2 − 𝑏2

| = (1 + 𝑎2 + 𝑏2)3

Solution:

𝐿𝐻𝑆 = |1 + 𝑎2 − 𝑏2 2𝑎𝑏 −2𝑏

2𝑎𝑏 1 − 𝑎2 + 𝑏2 2𝑎2𝑏 −2𝑎 1 − 𝑎2 − 𝑏2

|

=1

𝑎|1 + 𝑎2 − 𝑏2 2𝑎𝑏 −2𝑎𝑏

2𝑎𝑏 1 − 𝑎2 + 𝑏2 2𝑎2

2𝑏 −2𝑎 𝑎 − 𝑎3 − 𝑎𝑏2

| [Applying C3 → 𝑎𝐶3]

=1

𝑎|1 + 𝑎2 − 𝑏2 2𝑎𝑏 0

2𝑎𝑏 1 − 𝑎2 + 𝑏2 1 + 𝑎2 + 𝑏2

2𝑏 −2𝑎 −𝑎 − 𝑎3 − 𝑎𝑏2


=1 + 𝑎2 + 𝑏2

𝑎|1 + 𝑎2 − 𝑏2 2𝑎𝑏 0

2𝑎𝑏 1 − 𝑎2 + 𝑏2 12𝑏 −2𝑎 −𝑎

| [Taking 1 + a2 + 𝑏2 as common from 𝐶3]

=1+𝑎2+𝑏2

𝑎2 |1 + 𝑎2 − 𝑏2 2𝑎𝑏 0

2𝑎2𝑏 𝑎 − 𝑎3 + 𝑎𝑏2 𝑎2𝑏 −2𝑎 −𝑎

| [Applying 𝑅2 → 𝑎𝑅2]

=1+𝑎2+𝑏2

𝑎2 |1 + 𝑎2 − 𝑏2 2𝑎𝑏 02𝑎2𝑏 + 2𝑏 −𝑎 − 𝑎3 + 𝑎𝑏2 0

2𝑏 −2𝑎 −𝑎

| [Applying R2 → R2 + 𝑅3]

= (1 + 𝑎2 + 𝑏2) |1 + 𝑎2 − 𝑏2 2𝑏 02𝑎2𝑏 + 2𝑏 −1 − 𝑎2 + 𝑏2 0

2𝑏 −2 −1

| [Taking 𝑎 as common from C2 and C3]

= (1 + 𝑎2 + 𝑏2)(−1){(1 + 𝑎2 − 𝑏2)(−1 − 𝑎2 + 𝑏2) − 2𝑏(2𝑎2𝑏 + 2𝑏)} [Expanding along C3]

= −(1 + 𝑎2 + 𝑏2){−1 − 𝑎2 + 𝑏2 − 𝑎2 − 𝑎4 + 𝑎2𝑏2 + 𝑏2 + 𝑎2𝑏2 − 𝑏4 − 4𝑎2𝑏2 − 4𝑏2}

= (1 + 𝑎2 + 𝑏2){1 + 𝑎4 + 4 + 2𝑎2 + 2𝑎2𝑏2 + 2𝑏2}

= (1 + 𝑎2 + 𝑏2)(1 + 𝑎2 + 𝑏2)2 = (1 + 𝑎2 + 𝑏2)3 = 𝑅𝐻𝑆

Hence, LHS = RHS




|𝑎2 + 1 𝑎𝑏 𝑎𝑐

𝑎𝑏 𝑏2 + 1 𝑏𝑐𝑐𝑎 𝑐𝑏 𝑐2 + 1

| = 1 + 𝑎2 + 𝑏2 + 𝑐2

Solution:

𝐿𝐻𝑆 = |𝑎2 + 1 𝑎𝑏 𝑎𝑐

𝑎𝑏 𝑏2 + 1 𝑏𝑐𝑐𝑎 𝑐𝑏 𝑐2 + 1

|

=1

𝑎𝑏𝑐|𝑎2 + 𝑎 𝑎2𝑏 𝑎2𝑐𝑎𝑏2 𝑏3 + 𝑏 𝑏2𝑐𝑐2𝑎 𝑐2𝑏 𝑐3 + 𝑐

| [Applying R1 → 𝑎𝑅1, 𝑅3 → 𝑏𝑅3, 𝑅3 → 𝑐𝑅3]

=𝑎𝑏𝑐

𝑎𝑏𝑐|𝑎2 + 1 𝑎2 𝑎2

𝑏2 𝑏2 + 1 𝑏2

𝑐2 𝑐2 𝑐2 + 1

| [Taking a as common from C1, 𝑏 from C2 and c from C3]

= |1 + 𝑎2 + 𝑏2 + 𝑐2 1 + 𝑎2 + 𝑏2 + 𝑐2 1 + 𝑎2 + 𝑏2 + 𝑐2

𝑏2 𝑏2 + 1 𝑏2

𝑐2 𝑐2 𝑐2 + 1

| [By R1 → 𝑅1 + 𝑅2 + 𝑅3]

= (1 + 𝑎2 + 𝑏2 + 𝑐2) |1 1 1𝑏2 𝑏2 + 1 𝑏2

𝑐2 𝑐2 𝑐2 + 1| [Taking 1 + a2 + 𝑏2 + 𝑐2 as common from R1]

= (1 + 𝑎2 + 𝑏2 + 𝑐2) |0 0 1

−1 1 𝑏2

0 −1 𝑐2 + 1| [Applying C1 → 𝐶1 − 𝐶2, 𝐶2 → 𝐶2 − 𝐶3]

= (1 + 𝑎2 + 𝑏2 + 𝑐2){1 − 0} [Expanding along R1]

= 1 + 𝑎2 + 𝑏2 + 𝑐2 = 𝑅𝐻𝑆

Hence, LHS = RHS

15. Let 𝐴 be a square matrix of order 3 × 3, then |𝑘𝐴| is equal to:

(A) 𝑘|𝐴|

(B) 𝑘2|𝐴|

(C) 𝑘3|𝐴|

(D) 3𝑘|𝐴|



Solution:

If 𝐵 be a square matrix of order 𝑛 × 𝑛, then |𝑘𝐵| = 𝑘𝑛−1|𝐵|

Therefore, |𝑘𝐴| = 𝑘3−1|𝐴| = 𝑘2|𝐴|


16. Which of the following is correct

(A) Determinant is a square matrix

(B) Determinant is a number associated to a matrix

(C) Determinant is a number associated to a square matrix

(D) None of these

Solution:

Determinant is a number associated to a square matrix

Hence, the option (𝐶) is correct.

Exercise 𝟒. 𝟑

1. Find area of the triangle with vertices at the point given in each of the following:

(i) (1, 0), (6, 0), (4, 3)

(ii) (2, 7), (1, 1), (10, 8)

(iii) (−2,−3), (3, 2), (−1,−8)

Solution:

i) Given vertices of the triangle are (1, 0), (6, 0), (4, 3)

We know, area of triangle =1

2|

𝑥1 𝑦1 1𝑥2 𝑦2 1𝑥3 𝑦3 1

|

Area of triangle =1

2|1 0 16 0 14 3 1

|

=1

2[1(0 − 3) − 0(6 − 4) + 1(18 − 0)] =

1

2(15) = 7.5 square units

Given vertices of the triangle are (2, 7), (1, 1), (10, 8)




2|

𝑥1 𝑦1 1𝑥2 𝑦2 1𝑥3 𝑦3 1

|

Area of triangle =1

2|2 7 11 1 110 8 1

|

=1

2[2(1 − 8) − 7(1 − 10) + 1(8 − 10)] =

1

2(47) = 25.5 square units

Given vertices of the triangle are (−2,−3), (3, 2), (−1,−8)


2|

𝑥1 𝑦1 1𝑥2 𝑦2 1𝑥3 𝑦3 1

|

Area of triangle =1

2|−2 −3 13 2 1

−1 −8 1|

=1

2[−2(2 + 8) + 3(3 + 1) + 1(−24 + 2)] =

1

2(−30) = 15

Area of triangle = 15 square units

2. Show that points 𝐴(𝑎, 𝑏 + 𝑐), 𝐵(𝑏, 𝑐 + 𝑎), 𝐶(𝑐, 𝑎 + 𝑏) are collinear.

Solution:

If the points 𝐴(𝑎, 𝑏 + 𝑐), 𝐵(𝑏, 𝑐 + 𝑎) and 𝐶(𝑐, 𝑎 + 𝑏) are collinear, then the area of triangle

𝐴𝐵𝐶 will be zero.


2|

𝑥1 𝑦1 1𝑥2 𝑦2 1𝑥3 𝑦3 1

|

Area of triangle 𝐴𝐵𝐶 =1

2|𝑎 𝑏 + 𝑐 1𝑏 𝑐 + 𝑎 1𝑐 𝑎 + 𝑏 1

|

=1

2|𝑎 𝑎 + 𝑏 + 𝑐 1𝑏 𝑎 + 𝑏 + 𝑐 1𝑐 𝑎 + 𝑏 + 𝑐 1


=1

2(𝑎 + 𝑏 + 𝑐) |

𝑎 1 1𝑏 1 1𝑐 1 1

| [Taking 𝑎 + 𝑏 + 𝑐 as common from C2]

= 0 [∵ 𝐶1 = 𝐶3]

Hence, the points 𝐴(𝑎, 𝑏 + 𝑐), 𝐵(𝑏, 𝑐 + 𝑎) and 𝐶(𝑐, 𝑎 + 𝑏) are collinear.



3. Find values of 𝑘 if area of triangle is 4 square units and vertices are

(i) (𝑘, 0), (4, 0), (0, 2)

(ii) (−2, 0), (0, 4), (0, 𝑘)

Solution:

(i)Given vertices of the triangle are (𝑘, 0), (4, 0), (0, 2)


2|

𝑥1 𝑦1 1𝑥2 𝑦2 1𝑥3 𝑦3 1

|

Area of triangle =1

2|𝑘 0 14 0 10 2 1

|

=1

2[𝑘(0 − 2) − 0(4 − 0) + 1(8 − 0)] =

1

2(−2𝑘 + 8) = −𝑘 + 4

According to question, area of triangle 𝐴𝐵𝐶 = 4 square units

Therefore, |−𝑘 + 4| = 4 ⇒ −𝑘 + 4 = ±4

⇒ −𝑘 + 4 = 4 or −𝑘 + 4 = −4

⇒ 𝑘 = 0 or 𝑘 = 8

Hence, the value of 𝑘 are 0 and 8.

ii) Given vertices of the triangle are (−2, 0), (0, 4), (0, 𝑘)


2|

𝑥1 𝑦1 1𝑥2 𝑦2 1𝑥3 𝑦3 1

|

Now, area of triangle =1

2|−2 0 10 4 10 𝑘 1

|

=1

2[−2(4 − 𝑘) − 0(0 − 0) + 1(0 − 0)] =

1

2(−8 + 2𝑘) = −4 + 𝑘

According to question, Area of triangle 𝐴𝐵𝐶 = 4 square units

Therefore, |−4 + 𝑘| = 4 ⇒ −4 + 𝑘 = ±4

⇒ −4 + 𝑘 = 4 or −4 + 𝑘 = −4

⇒ 𝑘 = 8 or 𝑘 = 0

Hence, the value of 𝑘 are 0 and 8.

4. (i) Find equation of line joining (1, 2) and (3, 6) using determinants.



(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.

Solution:

(i) Let, 𝑃(𝑥, 𝑦) be any point lie on the line joining 𝐴(1, 2) and 𝐵(3, 6). Hence, the points 𝐴, 𝐵

and 𝑃 will be collinear and area of triangle 𝐴𝐵𝐶 will be zero.

Therefore, area of triangle 𝐴𝐵𝑃 =1

2|1 2 13 6 1𝑥 𝑦 1

| = 0

⇒1

2[1(6 − 𝑦) − 2(3 − 𝑥) + 1(3𝑦 − 6𝑥)] = 0

⇒ 6 − 𝑦 − 6 + 2𝑥 + 3𝑦 − 6𝑥 = 0

⇒ −4𝑥 + 2𝑦 = 0

⇒ 2𝑥 = 𝑦

Hence the equation of line joining the points (1, 2) and (3, 6) is 2𝑥 − 𝑦 = 0

(ii) Let, 𝑃(𝑥, 𝑦) be any point lie on the line joining 𝐴(3, 1) and 𝐵(9, 3). Hence, the points 𝐴, 𝐵

and 𝑃 will be collinear and area of triangle 𝐴𝐵𝐶 will be zero.

Therefore, area of triangle 𝐴𝐵𝑃 =1

2|3 1 19 3 1𝑥 𝑦 1

| = 0

⇒1

2[3(3 − 𝑦) − 1(9 − 𝑥) + 1(9𝑦 − 3𝑥)] = 0

⇒ 9 − 3𝑦 − 9 + 𝑥 + 9𝑦 − 3𝑥 = 0

⇒ −2𝑥 + 6𝑦 = 0

⇒ 𝑥 = 3𝑦

Hence the equation of line joining the points (3, 1) and (9, 3) is 𝑥 − 3𝑦 = 0

5. If area of triangle is 35 sq. units with vertices (2, −6), (5, 4) and (𝑘, 4). Then 𝑘 is

(A) 12

(B) −2

(C) −12,−2

(D) 12,−2



Solution:

Given vertices of the triangle are (2, −6), (5, 4), (𝑘, 4)


2|

𝑥1 𝑦1 1𝑥2 𝑦2 1𝑥3 𝑦3 1

|

Hence, area of triangle =1

2|2 −6 15 4 1𝑘 4 1

|

=1

2[2(4 − 4) + 6(5 − 𝑘) + 1(20 − 4𝑘)] =

1

2(30 − 6𝑘 + 20 − 4𝑘) = 25 − 5𝑘

According to question, area of triangle = 35 square units

Therefore, |25 − 5𝑘| = 35

⇒ 25 − 5𝑘 = ±35

⇒ 25 − 5𝑘 = 35 or 25 − 5𝑘 = −35

⇒ 𝑘 =−10

5= −2 or 𝑘 =

60

5= 12

Hence, the option (𝐷) is correct.

Exercise 𝟒. 𝟒

1. Write Minors and Cofactors of the elements of following determinants:

(i) |2 −40 3

|

(ii) |𝑎 𝑐𝑏 𝑑

|

Solution:

(i)

Given determinant is |2 −40 3

|

We know, the minor of element 𝑎𝑖𝑗 is 𝑀𝑖𝑗 and the cofactor is 𝐴𝑖𝑗 = (−1)𝑖+𝑗𝑀𝑖𝑗, therefore,

The minor of element 𝑎11 is 𝑀11 = 3 and the cofactor is 𝐴11 = (−1)1+1𝑀11 = 3


The minor of element 𝑎21 is 𝑀21 = −4 and the cofactor is 𝐴21 = (−1)2+1𝑀21 = 4




(ii)

Given determinant is |𝑎 𝑐𝑏 𝑑

|

We know, the minor of element 𝑎𝑖𝑗 is 𝑀𝑖𝑗 and the cofactor is 𝐴𝑖𝑗 = (−1)𝑖+𝑗𝑀𝑖𝑗, therefore,

The minor of element 𝑎11 is 𝑀11 = 𝑑 and the cofactor is 𝐴11 = (−1)1+1𝑀11 = 𝑑

The minor of element 𝑎12 is 𝑀12 = 𝑏 and the cofactor is 𝐴12 = (−1)1+2𝑀12 = −𝑏

The minor of element 𝑎21 is 𝑀21 = 𝑐 and the cofactor is 𝐴21 = (−1)2+1𝑀21 = −𝑐

The minor of element 𝑎22 is 𝑀22 = 𝑎 and the cofactor is 𝐴22 = (−1)2+2𝑀22 = 𝑎

2. Write Minors and Cofactors of the elements of following determinants:

(i) |1 0 00 1 00 0 1

|

(ii) |1 0 43 5 −10 1 2

|

Solution:

(i)

Given determinant is |1 0 00 1 00 0 1

|

We know, minor of an element 𝑎𝑖𝑗 of a determinant is the determinant obtained by

deleting its 𝑖𝑡ℎ row and 𝑗𝑡ℎ column in which element 𝑎𝑖𝑗 lies. Minor of an element 𝑎𝑖𝑗 is

denoted by 𝑀𝑖𝑗.

Hence,

𝑀11 = |1 00 1

| = 1 − 0 = 1

𝑀12 = |0 00 1

| = 0 − 0 = 0

𝑀13 = |0 10 0

| = 0 − 0 = 0

𝑀21 = |0 00 1

| = 0 − 0 = 0

𝑀22 = |1 00 1

| = 1 − 0 = 1



𝑀23 = |1 00 0

| = 0 − 0 = 0

𝑀31 = |0 01 0

| = 0 − 0 = 0

𝑀32 = |1 00 0

| = 0 − 0 = 0

𝑀33 = |1 00 1

| = 1 − 0 = 1

And we know, cofactor of an element 𝑎𝑖𝑗, is denoted by 𝐴𝑖𝑗 is defined by 𝐴𝑖𝑗 = (−1)𝑖+𝑗𝑀𝑖𝑗,

therefore

𝐴11 = (−1)1+1𝑀11 = 1 𝐴12 = (−1)1+2𝑀12 = 0 𝐴13 = (−1)1+3𝑀13 = 0

𝐴21 = (−1)2+1𝑀21 = 0 𝐴22 = (−1)2+2𝑀22 = 1 𝐴23 = (−1)2+3𝑀233 = 0

𝐴31 = (−1)3+1𝑀31 = 0 𝐴32 = (−1)3+2𝑀32 = 0 𝐴33 = (−1)3+3𝑀33 = 1

(ii)

Given determinant is |1 0 43 5 −10 1 2

|

We know, minor of an element 𝑎𝑖𝑗 of a determinant is the determinant obtained by

deleting its 𝑖𝑡ℎ row and 𝑗𝑡ℎ column in which element 𝑎𝑖𝑗 lies. Minor of an element 𝑎𝑖𝑗 is

denoted by 𝑀𝑖𝑗.

Hence,

𝑀11 = |5 −11 2

| = 10 + 1 = 11

𝑀12 = |3 −10 2

| = 6 − 0 = 6

𝑀13 = |3 50 1

| = 3 − 0 = 3

𝑀21 = |0 41 2

| = 0 − 4 = −4

𝑀22 = |1 40 2

| = 2 − 0 = 2

𝑀23 = |1 00 1

| = 1 − 0 = 1

𝑀31 = |0 45 −1

| = 0 − 20 = −20

𝑀32 = |1 43 −1

| = −1 − 12 = −13

𝑀33 = |1 03 5

| = 5 − 0 = 5

And we know, cofactor of an element 𝑎𝑖𝑗, is denoted by 𝐴𝑖𝑗 is defined by 𝐴𝑖𝑗 = (−1)𝑖+𝑗𝑀𝑖𝑗,

therefore



𝐴11 = (−1)1+1𝑀11 = 11 𝐴12 = (−1)1+2𝑀12 = −6 𝐴13 = (−1)1+3𝑀13 = 3

𝐴21 = (−1)2+1𝑀21 = 4 𝐴22 = (−1)2+2𝑀22 = 2 𝐴23 = (−1)2+3𝑀233 = −1

𝐴31 = (−1)3+1𝑀31 = −20 𝐴32 = (−1)3+2𝑀32 = 13 𝐴33 = (−1)3+3𝑀33 = 5

3. Using cofactors of elements of second row, evaluate ∆= |5 3 82 0 11 2 3

|.

Solution:

We know, ∆ = |5 3 82 0 11 2 3

| = 𝑎21𝐴21 + 𝑎22𝐴22 + 𝑎23𝐴23

Here, 𝑎21 = 2, 𝑎22 = 0, 𝑎23 = 1 and

𝐴21 = (−1)2+1 |3 82 3

| = −(9 − 16) = 7

𝐴22 = (−1)2+2 |5 81 3

| = 15 − 8 = 7

𝐴23 = (−1)2+3 |5 31 2

| = −(10 − 3) = −7

Therefore, ∆= |5 3 82 0 11 2 3

| = 2(7) + 0(7) + 1(−7) = 7

4. Using Cofactors of elements of third column, evaluate ∆= |

1 𝑥 𝑦𝑧1 𝑦 𝑧𝑥1 𝑧 𝑥𝑦

|.

Solution:

We know, ∆= |


| = 𝑎13𝐴13 + 𝑎23𝐴23 + 𝑎33𝐴33

Here, 𝑎13 = 𝑦𝑧, 𝑎23 = 𝑧𝑥, 𝑎33 = 𝑥𝑦 and

𝐴13 = (−1)1+3 |1 𝑦1 𝑧

| = 𝑧 − 𝑦

𝐴23 = (−1)2+3 |1 𝑥1 𝑧

| = −(𝑧 − 𝑥) = 𝑥 − 𝑧

𝐴33 = (−1)3+3 |1 𝑥1 𝑦

| = 𝑦 − 𝑥

Therefore, ∆= |


| = 𝑦𝑧(𝑧 − 𝑦) + 𝑧𝑥(𝑥 − 𝑧) + 𝑥𝑦(𝑦 − 𝑥)



= 𝑦𝑧2 − 𝑦2𝑧 + 𝑧𝑥2 − 𝑥𝑧2 + 𝑥𝑦2 − 𝑥2𝑦

= 𝑧𝑥2 − 𝑥2𝑦 − 𝑥𝑧2 + 𝑥𝑦2 + 𝑦𝑧2 − 𝑦2𝑧

= 𝑥2(𝑧 − 𝑦) − 𝑥(𝑧2 − 𝑦2) + 𝑦𝑧(𝑧 − 𝑦)

= (𝑧 − 𝑦)[𝑥2 − 𝑥(𝑧 + 𝑦) + 𝑦𝑧]

= (𝑧 − 𝑦)[𝑥2 − 𝑥𝑧 − 𝑥𝑦 + 𝑦𝑧]

= (𝑧 − 𝑦)[𝑥(𝑥 − 𝑧) − 𝑦(𝑥 − 𝑧)]

= (𝑥 − 𝑧)(𝑧 − 𝑦)(𝑥 − 𝑦)

= (𝑥 − 𝑦)(𝑦 − 𝑧)(𝑧 − 𝑥)

Hence, |


| = (𝑥 − 𝑦)(𝑦 − 𝑧)(𝑧 − 𝑥)

5. If ∆= |

𝑎11 𝑎12 𝑎13

𝑎21 𝑎22 𝑎23

𝑎31 𝑎32 𝑎33

| 𝐴𝑖𝑗 is Cofactors of 𝑎𝑖𝑗 , then value of ∆ is given by

(A) 𝑎11𝐴11 + 𝑎12𝐴32 + 𝑎13𝐴33

(B) 𝑎11𝐴11 + 𝑎12𝐴21 + 𝑎13𝐴31

(C) 𝑎21𝐴11 + 𝑎22𝐴12 + 𝑎23𝐴13

(D) 𝑎11𝐴11 + 𝑎21𝐴21 + 𝑎31𝐴31

Solution:

The value of |

𝑎11 𝑎12 𝑎13

𝑎21 𝑎22 𝑎23

𝑎31 𝑎32 𝑎33

| is given by: 𝑎11𝐴11 + 𝑎21𝐴21 + 𝑎31𝐴31


Exercise 𝟒. 𝟓

1. Find adjoint of the matrix [1 23 4

]

Solution:

Given matrix is [1 23 4

]



We know, the adjoint of a square matrix 𝐴 = [𝑎𝑖𝑗]𝑛×𝑛 is defined as the transpose of the

matrix [𝐴𝑖𝑗]𝑛×𝑛, where 𝐴𝑖𝑗 is the cofactor of the element 𝑎𝑖𝑗. Adjoint of the matrix 𝐴 is

denoted by 𝑎𝑑𝑗 𝐴.

Let 𝐴 = [1 23 4

]

therefore, 𝐴11 = 4, 𝐴12 = −3, 𝐴21 = −2, 𝐴22 = 1

Hence, adjoint of matrix 𝐴 = [𝐴11 𝐴21

𝐴12 𝐴22] = [

4 −2−3 1

]

2. Find adjoint of the matrix [1 −1 22 3 5

−2 0 1]

Solution:

Given matrix is [1 −1 22 3 5

−2 0 1]

We know, the adjoint of a square matrix 𝐴 = [𝑎𝑖𝑗]𝑛×𝑛 is defined as the transpose of the

matrix [𝐴𝑖𝑗]𝑛×𝑛, where 𝐴𝑖𝑗 is the cofactor of the element 𝑎𝑖𝑗. Adjoint of the matrix 𝐴 is

denoted by 𝑎𝑑𝑗 𝐴.

Let 𝐴 = [1 −1 22 3 5

−2 0 1], therefore

𝐴11 = 3 𝐴12 = −12 𝐴13 = 6𝐴21 = 1 𝐴22 = 5 𝐴23 = 2

𝐴31 = −11 𝐴32 = −1 𝐴33 = 5

Hence, adjoint of matrix 𝐴 = [

𝐴11 𝐴21 𝐴31

𝐴12 𝐴22 𝐴32

𝐴13 𝐴23 𝐴33

] = [3 1 −11

−12 5 −16 2 5

]

3. Verify 𝐴(𝑎𝑑𝑗 𝐴) = (𝑎𝑑𝑗 𝐴). 𝐴 = |𝐴|. 𝐼 for the matrix [2 3

−4 −6]

Solution:

Given matrix is [2 3

−4 −6]



Let 𝐴 = [2 3

−4 −6],

therefore, 𝐴11 = −6 𝐴12 = 4 𝐴21 = −3 𝐴22 = 2

|𝐴| = −12 + 12 = 0

𝑎𝑑𝑗 𝐴 = [𝐴11 𝐴21

𝐴12 𝐴22] = [

−6 −34 2

]

𝐴(𝑎𝑑𝑗 𝐴) = [2 3

−4 −6] [

−6 −34 2

] = [−12 + 12 −6 + 624 − 24 12 − 12

] = [0 00 0

]

(𝑎𝑑𝑗 𝐴). 𝐴 = [−6 −34 2

] [2 3

−4 −6] = [

−12 + 12 −18 + 188 − 8 12 − 12

] = [0 00 0

]

|𝐴|. 𝐼 = 0. [1 00 1

] = [0 00 0

]

Hence, 𝐴(𝑎𝑑𝑗 𝐴) = (𝑎𝑑𝑗 𝐴). 𝐴 = |𝐴|. 𝐼 = [0 00 0

]

4. Verify 𝐴(𝑎𝑑𝑗 𝐴) = (𝑎𝑑𝑗 𝐴). 𝐴 = |𝐴|. 𝐼 for the matrix [1 −1 23 0 −21 0 3

]

Solution:

Given matrix is [1 −1 23 0 −21 0 3

]

Here, 𝐴 = [1 −1 23 0 −21 0 3

], therefore, |𝐴| = 1(0 − 0) + 1(9 + 2) + 2(0 − 0) = 11

𝐴11 = 0 𝐴12 = −11 𝐴13 = 0𝐴21 = 3 𝐴22 = 1 𝐴23 = −1𝐴31 = 2 𝐴32 = 8 𝐴33 = 3

Adjoint of matrix 𝐴 = [

𝐴11 𝐴21 𝐴31

𝐴12 𝐴22 𝐴32

𝐴13 𝐴23 𝐴33

] = [0 3 2

−11 1 80 −1 3

]

𝐴(𝑎𝑑𝑗 𝐴) = [1 −1 23 0 −21 0 3

] [0 3 2

−11 1 80 −1 3

]

= [0 + 11 + 0 3 − 1 − 2 2 − 8 + 60 + 0 + 0 9 + 0 + 2 6 + 0 − 60 + 0 + 0 3 + 0 − 3 2 + 0 + 9

] = [11 0 00 11 00 0 11

]

Now, (𝑎𝑑𝑗 𝐴). 𝐴 = [0 3 2

−11 1 80 −1 3

] [1 −1 23 0 −21 0 3

]



= [0 + 9 + 2 0 + 0 + 0 0 − 6 + 6

−11 + 3 + 8 11 + 0 + 0 −22 − 2 + 240 − 3 + 3 0 + 0 + 0 0 + 2 + 9

] = [11 0 00 11 00 0 11

]

|𝐴|. 𝐼 = 11. [1 0 00 1 00 0 1

] = [11 0 00 11 00 0 11

]

Hence, 𝐴(𝑎𝑑𝑗 𝐴) = (𝑎𝑑𝑗 𝐴). 𝐴 = |𝐴|. 𝐼 = [11 0 00 11 00 0 11

]

5. Find the inverse of the matrix (if it exists)

[2 −24 3

]

Solution:

Given matrix is [2 −24 3

]

Let 𝐴 = [2 −24 3

]

Therefore, 𝐴11 = 3 𝐴12 = −4 𝐴21 = 2 𝐴22 = 2

And |𝐴| = 6 + 8 = 14 ≠ 0 ⇒ 𝐴−1 exists.

We know, 𝐴−1 =1

|𝐴|𝑎𝑑𝑗 𝐴 =

⇒ 𝐴−1 =1

|𝐴|[𝐴11 𝐴21

𝐴12 𝐴22]

=> 1

14[

3 2−4 2

]

Hence, the inverse of the matrix [2 −24 3

] is 1

14[

3 2−4 2

]


[−1 5−3 2

]

Solution:

Given matrix is [−1 5−3 2

]



Let 𝐴 = [−1 5−3 2

]

Therefore, 𝐴11 = 2 𝐴12 = 3 𝐴21 = −5 𝐴22 = −1

|𝐴| = −2 + 15 = 13 ≠ 0 ⇒ 𝐴−1 exists.

We know 𝐴−1 =1

|𝐴|𝑎𝑑𝑗 𝐴

⇒ 𝐴−1 =1

|𝐴|[𝐴11 𝐴21

𝐴12 𝐴22] =

1

13[2 −53 −1

]

Hence, the inverse of the matrix [−1 5−3 2

] is 1

13[2 −53 −1

]


[1 2 30 2 40 0 5

]

Solution:

Given matrix is [1 2 30 2 40 0 5

]

Let 𝐴 = [1 2 30 2 40 0 5

]

Therefore, |𝐴| = 1(10 − 0) − 2(0 − 0) + 3(0 − 0) = 10 ≠ 0 ⇒ 𝐴−1 exists.

𝐴11 = 10 𝐴12 = 0 𝐴13 = 0𝐴21 = −10 𝐴22 = 5 𝐴23 = 0𝐴31 = 2 𝐴32 = −4 𝐴33 = 2

We know 𝐴−1 =1


⇒ 𝐴−1 =1

|𝐴|[

𝐴11 𝐴21 𝐴31

𝐴12 𝐴22 𝐴32

𝐴13 𝐴23 𝐴33

]

=1

10[10 −10 20 5 −40 0 2

]

Hence, the inverse of the matrix [1 2 30 2 40 0 5

] is 1

10[10 −10 20 5 −40 0 2

]




[1 0 03 3 05 2 −1

]

Solution:

Given matrix is [1 0 03 3 05 2 −1

]

Let 𝐴 = [1 0 03 3 05 2 −1

],

Therefore, |𝐴| = 1(−3 − 0) − 0(−3 − 0) + 0(6 − 15) = −3 ≠ 0 ⇒ 𝐴−1 exists.

𝐴11 = −3 𝐴12 = 3 𝐴13 = −9𝐴21 = 0 𝐴22 = −1 𝐴23 = −2𝐴31 = 0 𝐴32 = 0 𝐴33 = 3



⇒ 𝐴−1 =1

|𝐴|[

𝐴11 𝐴21 𝐴31

𝐴12 𝐴22 𝐴32

𝐴13 𝐴23 𝐴33

]

=1

3[−3 0 03 −1 0

−9 −2 3]

Hence, the inverse of the matrix [1 0 03 3 05 2 −1

] is 1

3[−3 0 03 −1 0

−9 −2 3]


[2 1 34 −1 0

−7 2 1]

Solution:

Given matrix is [2 1 34 −1 0

−7 2 1]



Let 𝐴 = [2 1 34 −1 0

−7 2 1]

Therefore, |𝐴| = 2(−1 − 0) − 1(4 − 0) + 3(8 − 7) = −3 ≠ 0 ⇒ 𝐴−1 exists.

𝐴11 = −1 𝐴12 = −4 𝐴13 = 1𝐴21 = 5 𝐴22 = 23 𝐴23 = −11𝐴31 = 3 𝐴32 = 12 𝐴33 = −6



⇒ 𝐴−1 =1

|𝐴|[

𝐴11 𝐴21 𝐴31

𝐴12 𝐴22 𝐴32

𝐴13 𝐴23 𝐴33

]

=1

3[−1 5 3−4 23 121 −11 −6

]

Hence, the inverse of the matrix [2 1 34 −1 0

−7 2 1] is

1

3[−1 5 3−4 23 121 −11 −6

]


[1 −1 20 2 −33 −2 4

]

Solution:

Given matrix is [1 −1 20 2 −33 −2 4

]

Let 𝐴 = [1 −1 20 2 −33 −2 4

]

Therefore, [𝐴] = 1(8 − 6) + 1(0 + 9) + 2(0 − 6) = −1 ≠ 0 ⇒ 𝐴−1 exists.

𝐴11 = 2 𝐴12 = −9 𝐴13 = −6𝐴21 = 0 𝐴22 = −2 𝐴23 = −1

𝐴31 = −1 𝐴32 = 3 𝐴33 = 2



⇒ 𝐴−1 =1

|𝐴|[

𝐴11 𝐴21 𝐴31

𝐴12 𝐴22 𝐴32

𝐴13 𝐴23 𝐴33

]



=1

−1[

2 0 −1−9 −2 3−6 −1 2

] = [−2 0 19 2 −36 1 −2

]

Hence, the inverse of the matrix [1 −1 20 2 −33 −2 4

] is [−2 0 19 2 −36 1 −2

]


[1 0 00 cos𝛼 sin 𝛼0 sin𝛼 − cos𝛼

]

Solution:

Given matrix is [1 0 00 cos𝛼 sin𝛼0 sin𝛼 − cos𝛼

]

Let 𝐴 = [1 0 00 cos𝛼 sin𝛼0 sin𝛼 − cos𝛼

], therefore

Now, |𝐴| = 1(− cos2 𝛼 − sin2 𝛼) + 0(0 − 0) + 0(0 − 0) = −1 ≠ 0

⇒ 𝐴−1 exists. Therefore

𝐴11 = 1 𝐴12 = 0 𝐴13 = 0𝐴21 = 0 𝐴22 = −cos𝛼 𝐴23 = −sin𝛼𝐴31 = 0 𝐴32 = −sin𝛼 𝐴33 = cos𝛼



⇒ 𝐴−1 =1

|𝐴|[

𝐴11 𝐴21 𝐴31

𝐴12 𝐴22 𝐴32

𝐴13 𝐴23 𝐴33

]

=1

−1[1 0 00 − cos𝛼 − sin𝛼0 −sin𝛼 cos𝛼

]

⇒ 𝐴−1 = [−1 0 00 cos𝛼 sin𝛼0 sin𝛼 − cos𝛼

]

Hence, the inverse of the matrix [1 0 00 cos𝛼 sin𝛼0 sin𝛼 − cos𝛼

] is [−1 0 00 cos𝛼 sin𝛼0 sin𝛼 − cos𝛼

]

12. Let 𝐴 = [3 72 5

] and 𝐵 = [6 87 9

]. Verify that (𝐴𝐵)−1 = 𝐵−1𝐴−1.



Solution:

Given matrices are 𝐴 = [3 72 5

] and 𝐵 = [6 87 9

]

Here, 𝐴 = [3 72 5

], therefore, 𝐴11 = 5 𝐴12 = −2 𝐴21 = −7 𝐴22 = 3

Now, |𝐴| = 15 − 14 = 1 ≠ 0 ⇒ 𝐴−1 exists.



⇒ 𝐴−1 =1

|𝐴|[𝐴11 𝐴21

𝐴12 𝐴22]

=1

1[

5 −7−2 3

]

= [5 −7

−2 3]

and 𝐵 = [6 87 9

], therefore, 𝐵11 = 9 𝐵12 = −7 𝐵21 = −8 𝐵22 = 6

|𝐵| = 54 − 56 = −2 ≠ 0 ⇒ 𝐵−1 exists.

𝐵−1 =1

|𝐵|𝑎𝑑𝑗 𝐵

=1

|𝐵|[𝐵11 𝐵21

𝐵12 𝐵22]

=1

−2[

9 −8−7 6

]

= [−

9

24

7

2−3

]

Now, 𝐵−1𝐴−1 = [−9 2⁄ 47 2⁄ −3

] [5 −7

−2 3]

= [−

45

2− 8

63

2+ 12

35

2+ 6 −

49

2− 9

]

= [−

61

2

87

247

2−

67

2

] …(i)

And 𝐴𝐵 = [3 72 5

] [6 87 9

] = [18 + 49 24 + 6312 + 35 16 + 45

] = [67 8747 61

]

|𝐴𝐵| = 67 × 61 − 87 × 47 = 4087 − 4089 = −2 ≠ 0 ⇒ (𝐴𝐵)−1 exists.

𝐴𝐵11 = 61 𝐴𝐵12 = −47 𝐴𝐵21 = −87 𝐴𝐵22 = 67



(𝐴𝐵)−1 =1

|𝐴𝐵|𝑎𝑑𝑗 𝐴𝐵 =

1

|𝐴𝐵|[𝐴𝐵11 𝐴𝐵21

𝐴𝐵12 𝐴𝐵22] =

1

−2[

61 −87−47 67

] = [−

61

2

87

247

2−

67

2

] …(ii)

From (i) and (ii)

(𝐴𝐵)−1 = 𝐵−1𝐴−1

Hence verified.

13. If 𝐴 = [3 1

−1 2] show that 𝐴2 − 5𝐴 + 7𝐼 = 0. Hence, find 𝐴−1.

Solution:

Given matrix is 𝐴 = [3 1

−1 2]

𝐿𝐻𝑆 = 𝐴2 − 5𝐴 + 7𝐼 = 𝐴𝐴 − 5𝐴 + 7𝐼

= [3 1

−1 2] [

3 1−1 2

] − 5 [3 1

−1 2] + 7 [

1 00 1

]

= [9 − 1 3 + 2

−3 − 2 −1 + 4] − [

15 5−5 10

] + [7 00 7

]

= [8 − 15 + 7 5 − 5 + 0−5 + 5 + 0 3 − 10 + 7

] = [0 00 0

] = 𝑂 = 𝑅𝐻𝑆

⇒ 𝐴2 − 5𝐴 + 7𝐼 = 𝑂

⇒ 𝐴2 − 5𝐴 = −7𝐼

Multiplying by 𝐴−1(because |𝐴| ≠ 0)

⇒ 𝐴𝐴𝐴−1 − 5𝐴𝐴−1 = −7𝐼𝐴−1

⇒ 𝐴𝐼 − 5𝐼 = −7𝐴−1 [∵ 𝐴𝐴−1 = 𝐼]

⇒ 7𝐴−1 = 5𝐼 − 𝐴 = 5 [1 00 1

] − [3 1

−1 2] = [

5 00 5

] − [3 1

−1 2] = [

2 −11 3

]

⇒ 𝐴−1 =1

7[2 −11 3

]

14. For the matrix 𝐴 = [3 21 1

], find the numbers 𝑎 and 𝑏 such that 𝐴2 + 𝑎𝐴 + 𝑏𝐼 = 𝑂.

Solution:

Given matrix is 𝐴 = [3 21 1

]



And 𝐴2 + 𝑎𝐴 + 𝑏𝐼 = 𝑂

⇒ [3 21 1

] [3 21 1

] + 𝑎 [3 21 1

] + 𝑏 [1 00 1

] = [0 00 0

]

⇒ [9 + 2 6 + 23 + 1 2 + 1

] + [3𝑎 2𝑎𝑎 𝑎

] + [𝑏 00 𝑏

] = [0 00 0

]

⇒ [11 + 3𝑎 + 𝑏 8 + 2𝑎 + 04 + 𝑎 + 0 3 + 𝑎 + 𝑏

] = [0 00 0

]

⇒ 4 + 𝑎 = 0

⇒ 𝑎 = −4

and 3 + 𝑎 + 𝑏 = 0

⇒ 𝑏 = −3 − 𝑎 = −3 + 4

= 1

Hence, 𝑎 = −4, 𝑏 = 1

15. For the matrix 𝐴 = [1 1 11 2 −32 −1 3

]

show that 𝐴3 − 6𝐴2 + 5𝐴 + 11𝐼 = 0. Hence, find 𝐴−1.

Solution:

Given matrix is 𝐴 = [1 1 11 2 −32 −1 3

]

𝐴2 = 𝐴. 𝐴 = [1 1 11 2 −32 −1 3

] [1 1 11 2 −32 −1 3

]

= [1 + 1 + 2 1 + 2 − 1 1 − 3 + 31 + 2 − 6 1 + 4 + 3 1 − 6 − 92 − 1 + 6 2 − 2 − 3 2 + 3 + 9

] = [4 2 1

−3 8 −147 −3 14

]

𝐴3 = 𝐴2. 𝐴 = [4 2 1

−3 8 −147 −3 14

] [1 1 11 2 −32 −1 3

]

= [4 + 2 + 2 4 + 4 − 1 4 − 6 + 3

−3 + 8 − 28 −3 + 16 + 14 −3 − 24 − 427 − 3 + 28 7 − 6 − 14 7 + 9 + 42

] = [8 7 1

−23 27 −6932 −13 58

]

LHS = 𝐴3 − 6𝐴2 + 5𝐴 + 11𝐼

= [8 7 1

−23 27 −6932 −13 58

] − 6 [4 2 1

−3 8 −147 −3 14

] + 5 [1 1 11 2 −32 −1 3

] + 11 [1 0 00 1 00 0 1

]



= [8 7 1

−23 27 −6932 −13 58

] − [24 12 6

−18 48 −8442 −18 84

] + [5 5 55 10 −1510 −5 15

] + [11 0 00 11 00 0 11

]

= [8 − 24 + 5 + 11 7 − 12 + 5 + 0 1 − 6 + 5 + 0

−23 + 18 + 5 + 0 27 − 48 + 10 + 11 −69 + 84 − 15 + 032 − 42 + 10 + 0 −13 + 18 − 5 + 0 58 − 84 + 15 + 11

]

= [0 0 00 0 00 0 0

] = 𝑂 = 𝑅𝐻𝑆

Now, 𝐴3 − 6𝐴2 + 5𝐴 + 11 𝐼 = 𝑂

⇒ 𝐴3 − 6𝐴2 + 5𝐴 = −11 𝐼

Multiplying by 𝐴−1 (because |𝐴| ≠ 0)

⇒ 𝐴2𝐴𝐴−1 − 6𝐴𝐴𝐴−1 + 5𝐴𝐴−1 = −11𝐴−1

⇒ 𝐴2𝐼 − 6𝐴𝐼 + 5𝐼 = −11𝐴−1 [∵ 𝐴𝐴−1 = 𝐼]

⇒ 11𝐴−1 = −𝐴2 + 6𝐴 − 5𝐼

⇒ 11𝐴−1 = −[4 2 1

−3 8 −147 −3 14

] + 6 [1 1 11 2 −32 −1 3

] − 5 [1 0 00 1 00 0 1

]

⇒ 11𝐴−1 = [−4 −2 −13 −8 14

−7 3 −14] + [

6 6 66 12 −1812 −6 18

] − [5 0 00 5 00 0 5

]

⇒ 11𝐴−1 = [−4 + 6 − 5 −2 + 6 + 0 −1 + 6 + 03 + 6 − 0 −8 + 12 − 5 14 − 18 + 0

−7 + 12 + 0 3 − 6 + 0 −14 + 18 − 5] = [

−3 4 59 −1 −45 −3 −1

]

⇒ 𝐴−1 =1

11[−3 4 59 −1 −45 −3 −1

]

Hence, 𝐴−1 =1

11[−3 4 59 −1 −45 −3 −1

]

16. If 𝐴 = [2 −1 1

−1 2 −11 −1 2

]

verify that 𝐴3 − 6𝐴2 + 9𝐴 − 4𝐼 = 𝑂 and hence find 𝐴−1.



Solution:

𝐴2 = 𝐴. 𝐴 = [2 −1 1

−1 2 −11 −1 2

] [2 −1 1

−1 2 −11 −1 2

]

= [4 + 1 + 1 −2 − 2 − 1 2 + 1 + 2

−2 − 2 − 1 1 + 4 + 1 −1 − 2 − 22 + 1 + 2 −1 − 2 − 2 1 + 1 + 4

] = [6 −5 5

−5 6 −55 −5 6

]

𝐴3 = 𝐴2. 𝐴 = [6 −5 5

−5 6 −55 −5 6

] [2 −1 1

−1 2 −11 −1 2

]

= [12 + 5 + 5 −6 − 10 − 5 6 + 5 + 10

−10 − 6 − 5 5 + 12 + 5 −5 − 6 − 1010 + 5 + 6 −5 − 10 − 6 5 + 5 + 12

] = [22 −21 21

−21 22 −2121 −21 22

]

LHS = 𝐴3 − 6𝐴2 + 9𝐴 − 4𝐼

= [22 −21 21

−21 22 −2121 −21 22

] − 6 [6 −5 5

−5 6 −55 −5 6

] + 9 [2 −1 1

−1 2 −11 −1 2

] − 4 [1 0 00 1 00 0 1

]

= [22 −21 21

−21 22 −2121 −21 22

] − [36 −30 30

−30 36 −3030 −30 36

] + [18 −9 9−9 18 −99 −9 18

] − [4 0 00 4 00 0 4

]

= [22 − 36 + 18 − 4 −21 + 30 − 9 + 0 21 − 30 + 9 + 0−21 + 30 − 9 − 0 22 − 36 + 18 − 4 −21 + 30 − 9 + 021 − 30 + 9 − 0 −21 + 30 − 9 − 0 22 − 36 + 18 − 4

]

= [0 0 00 0 00 0 0

] = 𝑂 = 𝑅𝐻𝑆

⇒ 𝐴3 − 6𝐴2 + 9𝐴 − 4𝐼 = 𝑂 ⇒ 𝐴3 − 6𝐴2 + 9𝐴 = 4𝐼

Post multiplying by 𝐴−1(because |𝐴| ≠ 0)

𝐴2𝐴𝐴−1 − 6𝐴𝐴𝐴−1 + 9𝐴𝐴−1 = 4𝐼𝐴−1

⇒ 𝐴2𝐼 − 6𝐴𝐼 + 9𝐼 = 4𝐴−1 [Because 𝐴𝐴−1 = 𝐼]

⇒ 4𝐴−1 = 𝐴2 − 6𝐴 + 9𝐼

⇒ 4𝐴−1 = [6 −5 5

−5 6 −55 −5 6

] − 6 [2 −1 1

−1 2 −11 −1 2

] + 9 [1 0 00 1 00 0 1

]

⇒ 4𝐴−1 = [6 −5 5

−5 6 −55 −5 6

] − [12 −6 6−6 12 −66 −6 12

] + [9 0 00 9 00 0 9

]

⇒ 4𝐴−1 = [6 − 12 + 9 −5 + 6 + 0 5 − 6 + 0−5 + 6 + 0 6 − 12 + 9 −5 + 6 + 05 − 6 + 0 −5 + 6 + 0 6 − 12 + 9

] = [3 1 −11 3 1

−1 1 3]

⇒ 𝐴−1 =1

4[

3 1 −11 3 1

−1 1 3]



Hence, 𝐴−1 =1

4[

3 1 −11 3 1

−1 1 3]

17. Let 𝐴 be a non-singular square matrix of order 3 × 3. Then |𝑎𝑑𝑗 𝐴| is equal to:

(A) |𝐴|

(B) |𝐴|2

(C) |𝐴|3

(D) 3|𝐴|

Solution:

Given that 𝐴 be a non-singular square matrix of order 3 × 3.

We know that 𝑎𝑑𝑗 𝐴 = |𝐴|𝐼

⇒ (𝑎𝑑𝑗 𝐴)𝐴 = |𝐴| [1 0 00 1 00 0 1

]

⇒ |(𝑎𝑑𝑗 𝐴)𝐴| = |𝐴| |1 0 00 1 00 0 1

| = |

|𝐴| 0 00 |𝐴| 00 0 |𝐴|

| = |𝐴|3

⇒ |𝑎𝑑𝑗 𝐴| = |𝐴|2,


18. If 𝐴 is an invertible matrix of order 2, then det(𝐴−1) is equal to:

(A) det(𝐴)

(B) 1

det(𝐴)

(C) 1

(D) 0

Solution:

Given that the matrix 𝐴 is invertible, hence, 𝐴−1 =1


The order of matrix is 2, so, let 𝐴 = |𝑎 𝑏𝑐 𝑑

|



Therefore, |𝐴| = 𝑎𝑑 − 𝑏𝑐 and 𝑎𝑑𝑗 𝐴 = [𝑑 −𝑏−𝑐 𝑎

]

𝐴−1 =1

|𝐴|𝑎𝑑𝑗 𝐴 = 𝐴−1 =

1

|𝐴|[𝑑 −𝑏−𝑐 𝑎

] = [

𝑑

|𝐴|−

𝑏

|𝐴|

−𝑐

|𝐴|

𝑎

|𝐴|

]

det(𝐴−1) = |𝐴−1| = |

𝑑

|𝐴|−

𝑏

|𝐴|

−𝑐

|𝐴|

𝑎

|𝐴|

|

=1

|𝐴|2|𝑑 −𝑏−𝑐 𝑎

| =1

|𝐴|2(𝑎𝑑 − 𝑏𝑐) =

1

|𝐴|2|𝐴| =

1

|𝐴|


Exercise 𝟒. 𝟔

1. Examine the consistency of the system of equations

𝑥 + 2𝑦 = 2

2𝑥 + 3𝑦 = 3

Solution:

The given system of equations are 𝑥 + 2𝑦 = 2 and 2𝑥 + 3𝑦 = 3

This system of equations can be written as 𝐴𝑋 = 𝐵.

Where, 𝐴 = [1 22 3

] , 𝑋 = [𝑥𝑦] and 𝐵 = [

23]

Now, |𝐴| = 3 − 4 = −1 ≠ 0 ⇒ 𝐴 is non-singular and so 𝐴−1 exists.

Hence, the system of equations are consistent.


2𝑥 − 𝑦 = 5

𝑥 + 𝑦 = 4

Solution:

The given system of equations are 2𝑥 − 𝑦 = 5 and 𝑥 + 𝑦 = 4

This system of equations can be written as 𝐴𝑋 = 𝐵.



Where, 𝐴 = [2 −11 1

] , 𝑋 = [𝑥𝑦] and 𝐵 = [

54]

Now, |𝐴| = 2 + 1 = 3 ≠ 0 ⇒ 𝐴 is non-singular and so 𝐴−1 exists.



𝑥 + 3𝑦 = 5

2𝑥 + 6𝑦 = 8

Solution:

The given system of equations: 𝑥 + 3𝑦 = 5 and 2𝑥 + 6𝑦 = 8

This system of equations can be written as 𝐴𝑋 = 𝐵, where

𝐴 = [1 32 6

] , 𝑋 = [𝑥𝑦] and 𝐵 = [

58]

Now, |𝐴| = 6 − 6 = 0 ⇒ 𝐴 is a singular matrix and so 𝐴−1 does not exist.

Now, 𝑎𝑑𝑗 𝐴 = [6 −3

−2 1]

(𝑎𝑑𝑗 𝐴) 𝐵 = [6 −3

−2 1] [

58] = [

30 − 24−10 + 8

] = [6

−2] ≠ 𝑂

So, there is no solutions of the given system of equations.

Hence, the system of equations are inconsistent.


𝑥 + 𝑦 + 𝑧 = 1

2𝑥 + 3𝑦 + 2𝑧 = 2

𝑎𝑥 + 𝑎𝑦 + 2𝑎𝑧 = 4

Solution:

The given system of equations are 𝑥 + 𝑦 + 𝑧 = 1, 2𝑥 + 3𝑦 + 2𝑧 = 2 and 𝑎𝑥 + 𝑎𝑦 + 2𝑎𝑧 =4




𝐴 = [1 1 12 3 2𝑎 𝑎 2𝑎

] , 𝑋 = [𝑥𝑦𝑧] and 𝐵 = [

123]

Now, |𝐴| = 1(6𝑎 − 2𝑎) − 1(4𝑎 − 2𝑎) + 1(2𝑎 − 3𝑎) = 𝑎 ≠ 0

⇒ 𝐴 is non-singular and so 𝐴−1 exists.



3𝑥 − 𝑦 − 2𝑧 = 2

2𝑦 − 𝑧 = −1

3𝑥 − 5𝑦 = 3

Solution:

The given system of equations are 3𝑥 − 𝑦 − 2𝑧 = 2, 2𝑦 − 𝑧 = −1 and 3𝑥 − 5𝑦 = 3


𝐴 = [3 −1 −20 2 −13 −5 0

] , 𝑋 = [𝑥𝑦𝑧] and 𝐵 = [

2−13

]

|𝐴| = 3(0 − 5) + 1(0 + 3) − 2(0 − 6) = −15 + 3 + 12 = 0

⇒ 𝐴 is a singular matrix and so 𝐴−1 does not exists. Now,

𝐴11 = −5 𝐴12 = −3 𝐴13 = −6𝐴21 = 10 𝐴22 = 6 𝐴23 = 12𝐴31 = 5 𝐴32 = 3 𝐴33 = 6

Now, 𝑎𝑑𝑗 𝐴 = [−5 10 5−3 6 3−6 12 6

]

Also (𝑎𝑑𝑗 𝐴)𝐵 = [−5 10 5−3 6 3−6 12 6

] [2

−13

] = [−10 − 10 + 15

−6 − 6 + 9−12 − 12 + 16

] = [−5−3−6

] ≠ 𝑂

So, there is no solutions of the given system of equations.

Hence, the system of equations are inconsistent.


5𝑥 − 𝑦 + 4𝑧 = 5

2𝑥 + 3𝑦 + 5𝑧 = 2



5𝑥 − 2𝑦 + 6𝑧 = −1

Solution:

The given system of equations are 5𝑥 − 𝑦 + 4𝑧 = 5, 2𝑥 + 3𝑦 + 5𝑧 = 2 and 5𝑥 − 2𝑦 +

6𝑧 = −1


𝐴 = [5 −1 42 3 55 −2 6

] , 𝑋 = [𝑥𝑦𝑧] and 𝐵 = [

52

−1]

Now, |𝐴| = 5(18 + 10) + 1(12 − 25) + 4(−4 − 15) = 140 − 13 − 76 = 51 ≠ 0

⇒ 𝐴 is non-singular and so 𝐴−1 exists.


7. Solve system of linear equations, using matrix method

5𝑥 + 2𝑦 = 4

7𝑥 + 3𝑦 = 5

Solution:

The given system of equations are 5𝑥 + 2𝑦 = 4 and 7𝑥 + 3𝑦 = 5


𝐴 = [5 27 3

] , 𝑋 = [𝑥𝑦] and 𝐵 = [

45]

|𝐴| = 15 − 14 = 1 ≠ 0 ⇒ 𝐴 is non-singular and so 𝐴−1 exists.


Now, 𝐴11 = 3 𝐴12 = −7 𝐴21 = −2 𝐴22 = 5



⇒ 𝐴−1 =1

|𝐴|[𝐴11 𝐴21

𝐴12 𝐴22] =

1

1[

3 −2−7 5

]

Also 𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦] = [

3 −2−7 5

] [45]

⇒ [𝑥𝑦] = [

12 − 10−28 + 25

] ⇒ [𝑥𝑦] = [

2−3

] ⇒ 𝑥 = 2, 𝑦 = −3




2𝑥 − 𝑦 = −2

3𝑥 + 4𝑦 = 3

Solution:

The given system of equations are 2𝑥 − 𝑦 = −2 and 3𝑥 + 4𝑦 = 3


𝐴 = [2 −13 4

] , 𝑋 = [𝑥𝑦] and 𝐵 = [

−23

]

|𝐴| = 8 + 3 = 11 ≠ 0 ⇒ 𝐴 is non-singular and so 𝐴−1 exists. Now,


Now, 𝐴11 = 4 𝐴12 = −3 𝐴21 = 1 𝐴22 = 2

𝐴−1 =1


=1

|𝐴|[𝐴11 𝐴21

𝐴12 𝐴22] =

1

11[

4 1−3 2

]

𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦] =

1

11[

4 1−3 2

] [−23

]

⇒ [𝑥𝑦] =

1

11[−8 + 36 + 6

] ⇒ [𝑥𝑦] = [

−5

1112

11

] ⇒ 𝑥 = −5

11, 𝑦 =

12

11


4𝑥 − 3𝑦 = 3

3𝑥 − 5𝑦 = 7

Solution:

The given system of equations are 4𝑥 − 3𝑦 = 3 and 3𝑥 − 5𝑦 = 7


𝐴 = [4 −33 −5

] , 𝑋 = [𝑥𝑦] and 𝐵 = [

37]

|𝐴| = −20 + 9 = −11 ≠ 0 ⇒ 𝐴 is non-singular and so 𝐴−1 exists.


Now, 𝐴11 = −5 𝐴12 = −3 𝐴21 = 3 𝐴22 = 4



𝐴−1 =1


=1

|𝐴|[𝐴11 𝐴21

𝐴12 𝐴22] =

1

−11[−5 3−3 4

]

Now, 𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦] = −

1

11[−5 3−3 4

] [37]

⇒ [𝑥𝑦] = −

1

11[−15 + 21−9 + 28

] ⇒ [𝑥𝑦] = [

−6

11

−19

11

] ⇒ 𝑥 = −6

11, 𝑦 = −

19

11


5𝑥 + 2𝑦 = 3

3𝑥 + 2𝑦 = 5

Solution:

The given system of equations are 5𝑥 + 2𝑦 = 3 and 3𝑥 + 2𝑦 = 5


𝐴 = [5 23 2

] , 𝑋 = [𝑥𝑦] and 𝐵 = [

35]

|𝐴| = 10 − 6 = 4 ≠ 0 ⇒ 𝐴 is non-singular and so 𝐴−1 exists.


Now, 𝐴11 = 2 𝐴12 = −3 𝐴21 = −2 𝐴22 = 5

𝐴−1 =1


=1

|𝐴|[𝐴11 𝐴21

𝐴12 𝐴22] =

1

4[

2 −2−3 5

]

𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦] =

1

4[

2 −2−3 5

] [35]

⇒ [𝑥𝑦] =

1

4[

6 − 10−9 + 25

] ⇒ [𝑥𝑦] = [

−4

416

4

] ⇒ 𝑥 = −1, 𝑦 = 4


2𝑥 + 𝑦 + 𝑧 = 1



𝑥 − 2𝑦 − 𝑧 =3

2

3𝑦 − 5𝑧 = 9

Solution:

The given system of equations are 2𝑥 + 𝑦 + 𝑧 = 1, 𝑥 − 2𝑦 − 𝑧 =3

2 and 3𝑦 − 5𝑧 = 9


𝐴 = [2 1 11 −2 −10 3 −5

] , 𝑋 = [𝑥𝑦𝑧] and 𝐵 = [

13

2

9

]

|𝐴| = 2(10 + 3) − 1(−5 − 0) + 1(3 − 0) = 26 + 5 + 3 = 34 ≠ 0

⇒ 𝐴 is non-singular and so 𝐴−1 exists. Now,

𝐴11 = 13 𝐴12 = 5 𝐴13 = 3𝐴21 = 8 𝐴22 = −10 𝐴23 = −6𝐴31 = 1 𝐴32 = 3 𝐴33 = −5

𝐴−1 =1


=1

34[13 8 15 −10 33 −6 −5

]

Now, 𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦𝑧] =

1

34[13 8 15 −10 33 −6 −5

] [

13

2

9

]

⇒ [𝑥𝑦𝑧] =

1

34[13 + 12 + 95 − 15 + 273 − 9 − 45

] ⇒ [𝑥𝑦𝑧] =

1

34[

3417

−51] = [

11

2

−3

2

] ⇒ 𝑥 = 1, 𝑦 =1

2, 𝑧 = −

3

2


𝑥 − 𝑦 + 𝑧 = 4

2𝑥 + 𝑦 − 3𝑧 = 0

𝑥 + 𝑦 + 𝑧 = 2

Solution:

The given system of equations are 𝑥 − 𝑦 + 𝑧 = 4, 2𝑥 + 𝑦 − 3𝑧 = 0 and



𝑥 + 𝑦 + 𝑧 = 2


𝐴 = [1 −1 12 1 −31 1 1

] , 𝑋 = [𝑥𝑦𝑧] and 𝐵 = [

402]

|𝐴| = 1(1 + 3) + 1(2 + 3) + 1(2 − 1) = 4 + 5 + 1 = 10 ≠ 0


𝐴11 = 4 𝐴12 = −5 𝐴13 = 1𝐴21 = 2 𝐴22 = 0 𝐴23 = −2𝐴31 = 2 𝐴32 = 5 𝐴33 = 3

Now, 𝐴−1 =1


=1

10[

4 2 2−5 0 51 −2 3

]

Also 𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦𝑧] =

1

10[

4 2 2−5 0 51 −2 3

] [402]

⇒ [𝑥𝑦𝑧] =

1

10[

16 + 0 + 4−20 + 0 + 10

4 + 0 + 6] ⇒ [

𝑥𝑦𝑧] =

1

10[

20−1010

] = [2

−11

]

⇒ 𝑥 = 2, 𝑦 = −1, 𝑧 = 1


2𝑥 + 3𝑦 + 3𝑧 = 5

𝑥 − 2𝑦 + 𝑧 = −4

3𝑥 − 𝑦 − 2𝑧 = 3

Solution:

The given system of equations are 2𝑥 + 3𝑦 + 3𝑧 = 5, 𝑥 − 2𝑦 + 𝑧 = −4 and 3𝑥 − 𝑦 − 2𝑧 =

3


𝐴 = [2 3 31 −2 13 −1 −2

] , 𝑋 = [𝑥𝑦𝑧] and 𝐵 = [

5−43

]

|𝐴| = 2(4 + 1) − 3(−2 − 3) + 3(−1 + 6) = 10 + 15 + 15 = 40 ≠ 0




𝐴11 = 5 𝐴12 = 5 𝐴13 = 5𝐴21 = 3 𝐴22 = −13 𝐴23 = 11𝐴31 = 9 𝐴32 = 1 𝐴33 = −7

Now, 𝐴−1 =1


=1

40[5 3 95 −13 15 11 −7

]

Also, 𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦𝑧] =

1

40[5 3 95 −13 15 11 −7

] [5

−43

]

⇒ [𝑥𝑦𝑧] =

1

40[25 − 12 + 2725 + 52 + 325 − 44 − 21

] ⇒ [𝑥𝑦𝑧] =

1

40[

4080

−40] = [

12

−1]

⇒ 𝑥 = 1, 𝑦 = 2, 𝑧 = −1


𝑥 − 𝑦 + 2𝑧 = 7

3𝑥 + 4𝑦 − 5𝑧 = −5

2𝑥 − 𝑦 + 3𝑧 = 12

Solution:

The given system of equations are 𝑥 − 𝑦 + 2𝑧 = 7, 3𝑥 + 4𝑦 − 5𝑧 = −5 and 2𝑥 − 𝑦 + 3𝑧 =

12


𝐴 = [1 −1 23 4 −52 −1 3

] , 𝑋 = [𝑥𝑦𝑧] and 𝐵 = [

7−512

]

|𝐴| = 1(12 − 5) + 1(9 + 10) + 2(−3 − 8) = 7 + 19 − 22 = 4 ≠ 0


𝐴11 = 7 𝐴12 = −19 𝐴13 = −11𝐴21 = 1 𝐴22 = −1 𝐴23 = −1

𝐴31 = −3 𝐴32 = 11 𝐴33 = 7

Now, 𝐴−1 =1


=1

4[

7 1 −3−19 −1 11−11 −1 7

]



Also, 𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦𝑧] =

1

4[

7 1 −3−19 −1 11−11 −1 7

] [7

−512

]

⇒ [𝑥𝑦𝑧] =

1

4[

49 − 5 − 36−133 + 5 + 132−77 + 5 + 84

]

⇒ [𝑥𝑦𝑧] =

1

4[4412

] = [113]

⇒ 𝑥 = 1, 𝑦 = 1, 𝑧 = 3

15. If 𝐴 = [2 −3 53 2 −41 1 −2

], find 𝐴−1. Using 𝐴−1 solve the system of equations

2𝑥 − 3𝑦 + 5𝑧 = 11

3𝑥 + 2𝑦 − 4𝑧 = −5

𝑥 + 𝑦 − 2𝑧 = −3

Solution:

𝐴 = [2 −3 53 2 −41 1 −2

]

|𝐴| = 2(−4 + 4) + 3(−6 + 4) + 5(3 − 2) = 0 − 6 + 5 = −1 ≠ 0


𝐴11 = 0 𝐴12 = 2 𝐴13 = 1𝐴21 = −1 𝐴22 = −9 𝐴23 = −5𝐴31 = 2 𝐴32 = 23 𝐴33 = 13



=1

−1[0 −1 22 −9 231 −5 13

] = [0 1 −2

−2 9 −23−1 5 −13

]

The given system of equations:

2𝑥 − 3𝑦 + 5𝑧 = 11

3𝑥 + 2𝑦 − 4𝑧 = −5

𝑥 + 𝑦 − 2𝑧 = −3




𝐴 = [2 −3 53 2 −41 1 −2

] , 𝑋 = [𝑥𝑦𝑧] and 𝐵 = [

11−5−3

]

𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦𝑧] = [

0 1 −2−2 9 −23−1 5 −13

] [11−5−3

]

⇒ [𝑥𝑦𝑧] = [

0 − 5 + 6−22 − 45 + 69−11 − 25 + 39

] ⇒ [𝑥𝑦𝑧] = [

123]

⇒ 𝑥 = 1, 𝑦 = 2, 𝑧 = 3

16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. The cost of 2 kg onion, 4 kg wheat

and 6 kg rice is ₹90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹70. Find cost of

each item per kg by matrix method.

Solution:

Let the cost of 1 kg of onion = ₹𝑥,

Let the cost of 1 kg of wheat = ₹𝑦 and

Let the cost of 1 kg rice = ₹𝑧

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. So 4𝑥 + 3𝑦 + 2𝑧 = 60

The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. So 2𝑥 + 4𝑦 + 6𝑧 = 90 and

The cost of 6 kg onion, 2 kg wheat and 3 kg rice is ₹70. So 6𝑥 + 2𝑦 + 3𝑧 = 70

Therefore the system of equations are:

4𝑥 + 3𝑦 + 2𝑧 = 60

2𝑥 + 4𝑦 + 6𝑧 = 90

6𝑥 + 2𝑦 + 3𝑧 = 70


𝐴 = [4 3 22 4 66 2 3

] , 𝑋 = [𝑥𝑦𝑧] and 𝐵 = [

609070

]

|𝐴| = 4(12 − 12) − 3(6 − 36) + 2(4 − 24) = 0 + 90 − 40 = 50 ≠ 0


𝐴11 = 0 𝐴12 = 30 𝐴13 = −20𝐴21 = −5 𝐴22 = 0 𝐴23 = 10𝐴31 = 10 𝐴32 = −20 𝐴33 = 10



Now, 𝐴−1 =1


1

50[

0 −5 1030 0 −20

−20 10 10]

Also 𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦𝑧] =

1

50[

0 −5 1030 0 −20

−20 10 10] [

609070

]

⇒ [𝑥𝑦𝑧] =

1

50[

0 − 450 + 7001800 + 0 − 1400

−1200 + 900 + 700]

⇒ [𝑥𝑦𝑧] =

1

50[250400400

] = [588]

⇒ 𝑥 = 5, 𝑦 = 8, 𝑧 = 8

Hence,

the cost of 1 kg of onion is ₹7, the cost of 1 kg of wheat is ₹8 and the cost of 1 kg rice is ₹8.

Miscellaneous Exercise on 𝟒

1. Prove that the determinant |𝑥 sin𝜃 cos 𝜃

− sin𝜃 −𝑥 1cos 𝜃 1 𝑥

| is independent of 𝜃.

Solution:

Let the given determinant be ∆= |𝑥 sin 𝜃 cos 𝜃

− sin𝜃 −𝑥 1cos 𝜃 1 𝑥

|

= 𝑥(−𝑥2 − 1) − sin 𝜃(−𝑥 sin 𝜃 − cos 𝜃) + cos 𝜃(− sin𝜃 + 𝑥 cos𝜃)

= −𝑥3 − 𝑥 + 𝑥 sin2 𝜃 + sin 𝜃 cos 𝜃 − cos𝜃 sin𝜃 + 𝑥 cos2 𝜃

= −𝑥3 − 𝑥 + 𝑥(sin2 𝜃 + cos2 𝜃)

= −𝑥3 − 𝑥 + 𝑥 = −𝑥3, which is independent of 𝜃.

2. Without expanding the determinant, prove that |𝑎 𝑎2 𝑏𝑐𝑏 𝑏2 𝑐𝑎𝑐 𝑐2 𝑎𝑏

| = |1 𝑎2 𝑎3

1 𝑏2 𝑏3

1 𝑐2 𝑐3

|

Solution:

LHS = |𝑎 𝑎2 𝑏𝑐𝑏 𝑏2 𝑐𝑎𝑐 𝑐2 𝑎𝑏

|



=1

𝑎𝑏𝑐|𝑎2 𝑎3 𝑎𝑏𝑐𝑏2 𝑏3 𝑎𝑏𝑐𝑐2 𝑐3 𝑎𝑏𝑐

| [Applying 𝑅1 → 𝑎𝑅1, 𝑅2 → 𝑏𝑅2, 𝑅3 → 𝑐𝑅3]

=𝑎𝑏𝑐

𝑎𝑏𝑐|𝑎2 𝑎3 1𝑏2 𝑏3 1𝑐2 𝑐3 1

| [Taking 𝑎𝑏𝑐 as common from 𝐶3]

= (−1)1 |1 𝑎3 𝑎2

1 𝑏3 𝑏2

1 𝑐3 𝑐2

| [Interchanging 𝐶1 ⟷ 𝐶3]

= (−1)2 |1 𝑎2 𝑎3

1 𝑏2 𝑏3

1 𝑐2 𝑐3

| [Interchanging 𝐶2 ⟷ 𝐶3]

= |1 𝑎2 𝑎3

1 𝑏2 𝑏3

1 𝑐2 𝑐3

| = RHS

Hence proved.

3. Evaluate |

cos𝛼 cos𝛽 cos𝛼 sin𝛽 − sin𝛼− sin𝛽 cos𝛽 0

sin𝛼 cos𝛽 sin𝛼 sin𝛽 cos 𝛼|

Solution:

Given determinant is |

cos𝛼 cos𝛽 cos𝛼 sin𝛽 − sin𝛼− sin𝛽 cos𝛽 0

sin𝛼 cos𝛽 sin𝛼 sin𝛽 cos𝛼|

= −sin𝛼(− sin𝛼 sin2 𝛽 − sin𝛼 cos2 𝛽)

− 0(cos𝛼 cos𝛽 sin𝛼 sin𝛽 − cos𝛼 sin𝛽 sin𝛼 cos𝛽)

+cos𝛼(cos𝛼 cos2 𝛽 + cos𝛼 sin2 𝛽)

= sin2 𝛼(sin2 𝛽 + cos2 𝛽) + cos2 𝛼(cos2 𝛽 + sin2 𝛽) [Expanding along 𝐶3]

= sin2 𝛼(sin2 𝛽 + cos2 𝛽) + cos2 𝛼(cos2 𝛽 + sin2 𝛽)

= sin2 𝛼 + sin2 𝛼 = 1 [∵ sin2 𝜃 + cos2 𝜃 = 1]

4. If 𝑎, 𝑏 and 𝑐 are real numbers and

∆ = |𝑏 + 𝑐 𝑐 + 𝑎 𝑎 + 𝑏𝑐 + 𝑎 𝑎 + 𝑏 𝑏 + 𝑐𝑎 + 𝑏 𝑏 + 𝑐 𝑐 + 𝑎

| = 0

show that either 𝑎 + 𝑏 + 𝑐 = 0 or 𝑎 = 𝑏 = 𝑐.



Solution:

Given that ∆ = |𝑏 + 𝑐 𝑐 + 𝑎 𝑎 + 𝑏𝑐 + 𝑎 𝑎 + 𝑏 𝑏 + 𝑐𝑎 + 𝑏 𝑏 + 𝑐 𝑐 + 𝑎

| = 0

⇒ |2(𝑎 + 𝑏 + 𝑐) 2(𝑎 + 𝑏 + 𝑐) 2(𝑎 + 𝑏 + 𝑐)

𝑐 + 𝑎 𝑎 + 𝑏 𝑏 + 𝑐𝑎 + 𝑏 𝑏 + 𝑐 𝑐 + 𝑎

| = 0 [Applying 𝑅1 → 𝑅1 + 𝑅2 + 𝑅3]

⇒ 2(𝑎 + 𝑏 + 𝑐) |1 1 1

𝑐 + 𝑎 𝑎 + 𝑏 𝑏 + 𝑐𝑎 + 𝑏 𝑏 + 𝑐 𝑐 + 𝑎

| = 0 [Taking 2(𝑎 + 𝑏 + 𝑐) as common from 𝑅1][

⇒ 2(𝑎 + 𝑏 + 𝑐) |1 0 0

𝑐 + 𝑎 𝑏 − 𝑐 𝑏 − 𝑎𝑎 + 𝑏 𝑐 − 𝑎 𝑐 − 𝑏

| = 0 [Applying 𝐶2 → 𝐶2 − 𝐶1, 𝐶3 → 𝐶3 − 𝐶1]

⇒ 2(𝑎 + 𝑏 + 𝑐)1[(𝑏 − 𝑐)(𝑐 − 𝑏) − (𝑏 − 𝑎)(𝑐 − 𝑎)] = 0 [Expanding along 𝑅1]

⇒ 2(𝑎 + 𝑏 + 𝑐)[𝑏𝑐 − 𝑏2 − 𝑐2 + 𝑏𝑐 − (𝑏𝑐 − 𝑏𝑎 − 𝑎𝑐 + 𝑎2)] = 0

⇒ 2(𝑎 + 𝑏 + 𝑐)[𝑏𝑐 − 𝑏2 − 𝑐2 + 𝑎𝑏 + 𝑐𝑎 − 𝑎2] = 0

⇒ −(𝑎 + 𝑏 + 𝑐)[2𝑎2 + 2𝑏2 + 2𝑐2 − 2𝑎𝑏 − 2𝑏𝑐 − 2𝑐𝑎] = 0

⇒ −(𝑎 + 𝑏 + 𝑐)[(𝑎2 + 𝑏2 − 2𝑎𝑏) + (𝑏2 + 𝑐2 − 2𝑏𝑐) + (𝑐2 + 𝑎2 − 2𝑐𝑎)] = 0

⇒ −(𝑎 + 𝑏 + 𝑐)[(𝑎 − 𝑏)2 + (𝑏 − 𝑐)2 + (𝑐 − 𝑎)2] = 0

⇒ 𝑎 + 𝑏 + 𝑐 = 0 or (𝑎 − 𝑏)2 = 0, (𝑏 − 𝑐)2 = 0, (𝑐 − 𝑎)2 = 0

⇒ 𝑎 + 𝑏 + 𝑐 = 0 or 𝑎 − 𝑏 = 0, 𝑏 − 𝑐 = 0, 𝑐 − 𝑎 = 0

⇒ 𝑎 + 𝑏 + 𝑐 = 0 or a = b. 𝑏 = 𝑐, 𝑐 = 𝑎

Therefore, 𝑎 + 𝑏 + 𝑐 = 0 or 𝑎 = 𝑏 = 𝑐

5. Solve the equation |𝑥 + 𝑎 𝑥 𝑥

𝑥 𝑥 + 𝑎 𝑥𝑥 𝑥 𝑥 + 𝑎

| = 0, 𝑎 ≠ 0

Solution:

Given that |𝑥 + 𝑎 𝑥 𝑥


| = 0

⇒ |3𝑥 + 𝑎 3𝑥 + 𝑎 3𝑥 + 𝑎


| = 0 [Applying 𝑅1 → 𝑅1 + 𝑅2 + 𝑅3]



⇒ (3𝑥 + 𝑎) |1 1 1𝑥 𝑥 + 𝑎 𝑥𝑥 𝑥 𝑥 + 𝑎

| = 0 [Taking (3𝑥 + 𝑎) as common from 𝑅1]

⇒ (3𝑥 + 𝑎) |1 0 0𝑥 𝑎 0𝑥 0 𝑎

| = 0 [Applying 𝐶2 → 𝐶2 − 𝐶1, 𝐶3 → 𝐶3 + 𝐶1]

⇒ (3𝑥 + 𝑎)1[𝑎2 − 0] = 0 [Expanding along 𝑅1]

⇒ 𝑎2(3𝑥 + 𝑎) = 0

⇒ (3𝑥 + 𝑎) = 0 [∵ 𝑎 ≠ 0]

⇒ 𝑥 = −𝑎

3

Value of x is −𝑎

3

6. Prove that |𝑎2 𝑏𝑐 𝑎𝑐 + 𝑐2

𝑎2 + 𝑎𝑏 𝑏2 𝑎𝑐𝑎𝑏 𝑏2 + 𝑏𝑐 𝑐2

| = 4𝑎2𝑏2𝑐2

Solution:

LHS = |𝑎2 𝑏𝑐 𝑎𝑐 + 𝑐2


|

= |2𝑎2 0 2𝑎𝑐


| [Applying 𝑅1 → 𝑅1 + 𝑅2 − 𝑅3]

= 𝑎𝑏𝑐 |2𝑎 0 2𝑎

𝑎 + 𝑏 𝑏 𝑎𝑏 𝑏 + 𝑐 𝑐

| [Taking 𝑎, 𝑏, 𝑐 as common from 𝐶1, 𝐶2, 𝐶3]

= 𝑎𝑏𝑐 |2𝑎 0 0

𝑎 + 𝑏 𝑏 −𝑏𝑏 𝑏 + 𝑐 𝑐 − 𝑏

| [Applying 𝐶3 → 𝐶3 − 𝐶1]

= (𝑎𝑏𝑐)2𝑎[𝑏(𝑐 − 𝑏) − (−𝑏)(𝑏 + 𝑐)] [Expanding along 𝑅1]

= 2𝑎2𝑏𝑐[𝑏𝑐 − 𝑏2 + 𝑏2 + 𝑏𝑐] = 2𝑎2𝑏𝑐[2𝑏𝑐]

= 4𝑎2𝑏2𝑐2 = RHS

Hence proved

7. If 𝐴−1 = [3 −1 1

−15 6 −55 −2 2

] and 𝐵 = [1 2 −2

−1 3 00 −2 1

], find (𝐴𝐵)−1.



Solution:

Given that 𝐴−1 = [3 −1 1

−15 6 −55 −2 2

] and 𝐵 = [1 2 −2

−1 3 00 −2 1

]

Here, 𝐵 = [1 2 −2

−1 3 00 −2 1

]

Therefore, |𝐵| = 1(3 − 0) − 2(−1 − 0) − 2(2 − 0) = 1 ≠ 0 ⇒ 𝐵−1 exists.

𝐵11 = 3 𝐵12 = 1 𝐵13 = 2𝐵21 = 2 𝐵22 = 1 𝐵23 = 2𝐵31 = 6 𝐵32 = 2 𝐵33 = 5

We know, 𝐵−1 =1

|𝐵|𝑎𝑑𝑗 𝐵

=1

1[

𝐵11 𝐵21 𝐵31

𝐵12 𝐵22 𝐵32

𝐵13 𝐵23 𝐵33

] = [3 2 61 1 22 2 5

]

Also (𝐴𝐵)−1 = 𝐵−1𝐴−1, therefore

(𝐴𝐵)−1 = 𝐵−1𝐴−1𝑧

= [3 2 61 1 22 2 5

] [3 −1 1

−15 6 −55 −2 2

]

= [9 − 30 + 30 −3 + 12 − 12 3 − 10 + 123 − 15 + 10 −1 + 6 − 4 1 − 5 + 46 − 30 + 25 −2 + 12 − 10 2 − 10 + 10

] = [9 −3 5

−2 1 01 0 2

]

Hence, (𝐴𝐵)−1 = [9 −3 5

−2 1 01 0 2

]

8. Let 𝐴 = [1 −2 1

−2 3 11 1 5

]. Verify that

(i) (𝑎𝑑𝑗 𝐴)−1 = 𝑎𝑑𝑗(𝐴−1)

(ii) (𝐴−1)−1 = 𝐴

Solution:

(i)

Given matrix is 𝐴 = [1 −2 1

−2 3 11 1 5

], therefore

|𝐴| = 1(15 − 1) + 2(−10 − 1) + 1(−2 − 3) = −13 ≠ 0 ⇒ 𝐴−1 exists.



𝐴11 = 14 𝐴12 = 11 𝐴13 = −5𝐴21 = 11 𝐴22 = 4 𝐴23 = −3𝐴31 = −5 𝐴32 = −3 𝐴33 = −1

𝑎𝑑𝑗 𝐴 = [

𝐴11 𝐴21 𝐴31

𝐴12 𝐴22 𝐴32

𝐴13 𝐴23 𝐴33

] = [14 11 −511 4 −3−5 −3 −1

]

𝐴−1 =1


1

|𝐴|[

𝐴11 𝐴21 𝐴31

𝐴12 𝐴22 𝐴32

𝐴13 𝐴23 𝐴33

] =1

−13[14 11 −511 4 −3−5 −3 −1

] …(i)

Let, 𝐵 = 𝑎𝑑𝑗 𝐴, so, 𝐵 = [14 11 −511 4 −3−5 −3 −1

], therefore

|𝐵| = 14(−4 − 9) − 11(−11 − 15) − 5(−33 + 20) = −182 + 286 + 65 = 169 ≠ 0

⇒ 𝐵−1 exists.

𝐵11 = −13 𝐵12 = 26 𝐵13 = −13𝐵21 = 26 𝐵22 = −39 𝐵23 = −13

𝐵31 = −13 𝐵32 = −13 𝐵33 = −65

𝐵−1 =1

|𝐵|[

𝐵11 𝐵21 𝐵31

𝐵12 𝐵22 𝐵32

𝐵13 𝐵23 𝐵33

] =1

169[−13 26 −1326 −39 −13

−13 −13 −65] =

1

13[−1 2 −12 −3 −1

−1 −1 −5]

⇒ (𝑎𝑑𝑗 𝐴)−1 =1

13[−1 2 −12 −3 −1

−1 −1 −5] …(ii)

Let, 𝐶 = 𝐴−1, so, 𝐶 =1

−13[14 11 −511 4 −3−5 −3 −1

] =

[ −

14

13−

11

13

5

13

−11

13−

4

13

3

135

13

3

13

1

13]

, therefore

𝐶11 = −1

13𝐶12 =

2

13𝐶13 = −

1

13

𝐶21 =2

13𝐶22 = −

3

13𝐶23 = −

1

13

𝐶31 = −1

13𝐶32 = −

1

13𝐶33 = −

5

13

𝐴𝑑𝑗 𝐶 = [

𝐶11 𝐶21 𝐶31

𝐶12 𝐶22 𝐶32

𝐶13 𝐶23 𝐶33

] =

[ −

1

13

2

13−

1

132

13−

3

13−

1

13

−1

13−

1

13−

5

13]

=1

13[−1 2 −12 −3 −1

−1 −1 −5]

⇒ 𝐴𝑑𝑗 𝐶 = 𝑎𝑑𝑗 (𝐴−1) =1

13[−1 2 −12 −3 −1

−1 −1 −5] …(iii)

From the equations (ii) and (iii) we have, (𝑎𝑑𝑗 𝐴)−1 = 𝑎𝑑𝑗(𝐴−1)



(ii) From the equation (i), we have,

𝐴−1 =1

−13[14 11 −511 4 −3−5 −3 −1

]

Let, 𝐷 = 𝐴−1, so, 𝐷 =1

−13[14 11 −511 4 −3−5 −3 −1

] =

[ −

14

13−

11

13

5

13

−11

13−

4

13

3

135

13

3

13

1

13]

, therefore

|𝐷| = −(1

13)3

[14(−4 − 9) − 11(−11 − 15) − 5(−33 + 20)]

= −(1

13)3(169) = −

1

13≠ 0 ⇒ 𝐷−1 exists.

𝐷11 = −1

13𝐷12 =

2

13𝐷13 = −

1

13

𝐷21 =2

13𝐷22 = −

3

13𝐷23 = −

1

13

𝐷31 = −1

13𝐷32 = −

1

13𝐷33 = −

5

13

𝐷−1 =1

|𝐷|[

𝐷11 𝐷21 𝐷31

𝐷12 𝐷22 𝐷32

𝐷13 𝐷23 𝐷33

] =1

−1 13⁄

[ −

1

13

2

13−

1

132

13−

3

13−

1

13

−1

13−

1

13−

5

13]

= [1 −2 1

−2 3 11 1 5

]

⇒ 𝐷−1 = (𝐴−1)−1 = [1 −2 1

−2 3 11 1 5

] = 𝐴

9. Evaluate |

𝑥 𝑦 𝑥 + 𝑦𝑦 𝑥 + 𝑦 𝑥

𝑥 + 𝑦 𝑥 𝑦|.

Solution:



𝑥 + 𝑦 𝑥 𝑦|

= |

2(𝑥 + 𝑦) 𝑦 𝑥 + 𝑦

2(𝑥 + 𝑦) 𝑥 + 𝑦 𝑥

2(𝑥 + 𝑦) 𝑥 𝑦| [Applying 𝐶1 → 𝐶1 + 𝐶2 + 𝐶3]

= 2(𝑥 + 𝑦) |

1 𝑦 𝑥 + 𝑦1 𝑥 + 𝑦 𝑥1 𝑥 𝑦

| [Taking 2(𝑥 + 𝑦) as common from 𝐶1]



= 2(𝑥 + 𝑦) |

0 −𝑥 𝑦0 𝑦 𝑥 − 𝑦1 𝑦 𝑦 + 𝑘

| [Applying 𝑅1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]

= 2(𝑥 + 𝑦){(−𝑥)(𝑥 − 𝑦) − 𝑦. 𝑦} [Expanding along 𝐶1]

= 2(𝑥 + 𝑦)(−𝑥2 + 𝑥𝑦 − 𝑦2)

= −2(𝑥 + 𝑦)(𝑥2 − 𝑥𝑦 + 𝑦2) = −2(𝑥3 + 𝑦3)

Hence, |


𝑥 + 𝑦 𝑥 𝑦| = −2(𝑥3 + 𝑦3)

10. Evaluate |

1 𝑥 𝑦1 𝑥 + 𝑦 𝑦1 𝑥 𝑥 + 𝑦

|

Solution:


1 𝑥 𝑦1 𝑥 + 𝑦 𝑦1 𝑥 𝑥 + 𝑦

|

= |

0 −𝑦 00 𝑦 −𝑥1 𝑥 𝑥 + 𝑦

| [Applying 𝑅1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]

= {(−𝑦)(−𝑥) − 𝑦. 0} [Expanding along 𝐶1]

= 𝑥𝑦

Hence, |

1 𝑥 𝑦1 𝑥 + 𝑦 𝑦1 𝑥 𝑥 + 𝑦

| = 𝑥𝑦

11. Using properties of determinants, prove that:

|

𝛼 𝛼2 𝛽 + 𝛾

𝛽 𝛽2 𝛾 + 𝛼

𝛾 𝛾2 𝛼 + 𝛽

| = (𝛽 − 𝛾)(𝛾 − 𝛼)(𝛼 − 𝛽)(𝛼 + 𝛽 + 𝛾)

Solution:

LHS = |

𝛼 𝛼2 𝛽 + 𝛾

𝛽 𝛽2 𝛾 + 𝛼

𝛾 𝛾2 𝛼 + 𝛽

|



= |

𝛼 𝛼2 𝛼 + 𝛽 + 𝛾

𝛽 𝛽2 𝛼 + 𝛽 + 𝛾

𝛾 𝛾2 𝛼 + 𝛽 + 𝛾

| [Applying 𝐶3 → 𝐶1 + 𝐶3]

= (𝛼 + 𝛽 + 𝛾) |𝛼 𝛼2 1𝛽 𝛽2 1

𝛾 𝛾2 1

| [Taking α + β + γ as common from 𝐶3]

= (𝛼 + 𝛽 + 𝛾) |

𝛼 − 𝛽 𝛼2 − 𝛽2 0

𝛽 − 𝛾 𝛽2 − 𝛾2 0

𝛾 𝛾2 1

| [Applying 𝑅1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]

= (𝛼 + 𝛽 + 𝛾)(𝛼 − 𝛽)(𝛽 − 𝛾) |

1 𝛼 + 𝛽 01 𝛽 + 𝛾 0

𝛾 𝛾2 1|

[Taking 𝛼 − 𝛽 as common from 𝑅1 and 𝛽 − 𝛾 as common from 𝑅2]

= (𝛼 + 𝛽 + 𝛾)(𝛼 − 𝛽)(𝛽 − 𝛾){(𝛽 + 𝛾) − (𝛼 + 𝛽)} [Expanding along 𝐶3]

= (𝛼 + 𝛽 + 𝛾)(𝛼 − 𝛽)(𝛽 − 𝛾)(𝛾 − 𝛼) = RHS

Hence proved.


|

𝑥 𝑥2 1 + 𝑝𝑥3

𝑦 𝑦2 1 + 𝑝𝑦3

𝑧 𝑧2 1 + 𝑝𝑧3

| = (1 + 𝑝𝑥𝑦𝑧)(𝑥 − 𝑦)(𝑦 − 𝑧)(𝑧 − 𝑥), where p is any scalar.

Solution:

LHS = |

𝑥 𝑥2 1 + 𝑝𝑥3

𝑦 𝑦2 1 + 𝑝𝑦3

𝑧 𝑧2 1 + 𝑝𝑧3

|

= |𝑥 𝑥2 1𝑦 𝑦2 1

𝑧 𝑧2 1

| + |

𝑥 𝑥2 𝑝𝑥3

𝑦 𝑦2 𝑝𝑦3

𝑧 𝑧2 𝑝𝑧3

|

= (−1)1 |1 𝑥2 𝑥1 𝑦2 𝑦

1 𝑧2 𝑧

| + 𝑝 |𝑥 𝑥2 𝑥3

𝑦 𝑦2 𝑦3

𝑧 𝑧2 𝑧3

| [Taking 𝑝 as common from 𝐶3]

= (−1)2 |1 𝑥 𝑥2

1 𝑦 𝑦2

1 𝑧 𝑧2

| + 𝑝𝑥𝑦𝑧 |1 𝑥 𝑥2

1 𝑦 𝑦2

1 𝑧 𝑧2

| [Taking 𝑥, 𝑦, 𝑧 as common from 𝑅1, 𝑅2, 𝑅3 respectively]



= (1 + 𝑝𝑥𝑦𝑧) |1 𝑥 𝑥2

1 𝑦 𝑦2

1 𝑧 𝑧2

| [Taking |1 𝑥 𝑥2

1 𝑦 𝑦2

1 𝑧 𝑧2

| as common]

= (1 + 𝑝𝑥𝑦𝑧) |0 𝑥 − 𝑦 𝑥2 − 𝑦2

0 𝑦 − 𝑧 𝑦2 − 𝑧2

1 𝑧 𝑧2

| [Applying 𝑅1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]

= (1 + 𝑝𝑥𝑦𝑧)(𝑥 − 𝑦)(𝑦 − 𝑧) |0 1 𝑥 + 𝑦0 1 𝑦 + 𝑧

1 𝑧 𝑧2

| [Taking 𝑥 − 𝑦 as common from 𝑅1 and 𝑦 − 𝑧 as common form 𝑅2]

= (1 + 𝑝𝑥𝑦𝑧)(𝑥 − 𝑦)(𝑦 − 𝑧){(𝑦 + 𝑧) − (𝑥 + 𝑦)} [Expanding along 𝐶1]

= (1 + 𝑝𝑥𝑦𝑧)(𝑥 − 𝑦)(𝑦 − 𝑧)(𝑧 − 𝑥) = RHS

Hence proved.


|3𝑎 −𝑎 + 𝑏 −𝑎 + 𝑐

−𝑏 + 𝑎 3𝑏 −𝑏 + 𝑐−𝑐 + 𝑎 −𝑐 + 𝑏 3𝑐

| = 3(𝑎 + 𝑏 + 𝑐)(𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎)

Solution:

LHS = |3𝑎 −𝑎 + 𝑏 −𝑎 + 𝑐

−𝑏 + 𝑎 3𝑏 −𝑏 + 𝑐−𝑐 + 𝑎 −𝑐 + 𝑏 3𝑐

|

= |𝑎 + 𝑏 + 𝑐 −𝑎 + 𝑏 −𝑎 + 𝑐𝑎 + 𝑏 + 𝑐 3𝑏 −𝑏 + 𝑐𝑎 + 𝑏 + 𝑐 −𝑐 + 𝑏 3𝑐


= (𝑎 + 𝑏 + 𝑐) |1 −𝑎 + 𝑏 −𝑎 + 𝑐1 3𝑏 −𝑏 + 𝑐1 −𝑐 + 𝑏 3𝑐

| [Taking (𝑎 + 𝑏 + 𝑐) as common from 𝐶1]

= (𝑎 + 𝑏 + 𝑐) |0 −𝑎 − 2𝑏 −𝑎 + 𝑏0 2𝑏 + 𝑐 −𝑏 − 2𝑐1 −𝑐 + 𝑏 3𝑐

| [Applying 𝑅1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]

= (𝑎 + 𝑏 + 𝑐){(−𝑎 − 2𝑏)(−𝑏 − 2𝑐) − (2𝑏 + 𝑐)(−𝑎 + 𝑏)} [Expanding along 𝐶1]

= (𝑎 + 𝑏 + 𝑐)(𝑎𝑏 + 2𝑎𝑐 + 2𝑏2 + 4𝑏𝑐 − (−2𝑎𝑏 + 2𝑏2 − 𝑎𝑐 + 𝑏𝑐))

= (𝑎 + 𝑏 + 𝑐)(3𝑎𝑏 + 3𝑏𝑐 + 3𝑐𝑎) = 3(𝑎 + 𝑏 + 𝑐)(𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎) = RHS

Hence proved.




|

1 1 + 𝑝 1 + 𝑝 + 𝑞2 3 + 2𝑝 4 + 3𝑝 + 2𝑞3 6 + 3𝑝 10 + 6𝑝 + 3𝑞

| = 1

Solution:

LHS = |

1 1 + 𝑝 1 + 𝑝 + 𝑞2 3 + 2𝑝 4 + 3𝑝 + 2𝑞3 6 + 3𝑝 10 + 6𝑝 + 3𝑞

|

= |

1 1 + 𝑝 1 + 𝑝 + 𝑞0 1 2 + 𝑝0 3 7 + 3𝑝

| [Applying 𝑅2 → 𝑅2 − 2𝑅1, 𝑅3 → 𝑅3 − 3𝑅1]

= 1{1. (7 + 3𝑝) − (3)(2 + 𝑝)} [Expanding along 𝐶1]

= 7 + 3𝑝 − 6 − 3𝑝 = 1 = RHS

Hence proved.


|

sin 𝛼 cos𝛼 cos(𝛼 + 𝛿)

sin𝛽 cos𝛽 cos(𝛽 + 𝛿)

sin𝛾 cos 𝛾 cos(𝛾 + 𝛿)| = 0

Solution:

LHS = |

sin𝛼 cos 𝛼 cos(𝛼 + 𝛿)

sin𝛽 cos 𝛽 cos(𝛽 + 𝛿)

sin𝛾 cos 𝛾 cos(𝛾 + 𝛿)|

= |

sin𝛼 cos𝛼 cos𝛿 − sin𝛼 sin𝛿 cos(𝛼 + 𝛿)

sin𝛽 cos𝛽 cos𝛿 − sin𝛽 sin 𝛿 cos(𝛽 + 𝛿)

sin𝛾 cos 𝛾 cos 𝛿 − sin 𝛾 sin𝛿 cos(𝛾 + 𝛿)| [Applying 𝐶2 → cos𝛿𝐶2 − sin 𝛿 𝐶1]

= |

sin𝛼 cos(𝛼 + 𝛿) cos(𝛼 + 𝛿)

sin𝛽 cos(𝛽 + 𝛿) cos(𝛽 + 𝛿)

sin𝛾 cos(𝛾 + 𝛿) cos(𝛾 + 𝛿)|

= 0 = RHS [∵ 𝐶2 = 𝐶3]

Hence proved.



16. Solve the system of equations:

2

𝑥+

3

𝑦+

10

𝑧= 4

4

𝑥−

6

𝑦+

5

𝑧= 1

6

𝑥+

9

𝑦−

20

𝑧= 2

Solution:

The given system of equations are

2

𝑥+

3

𝑦+

10

𝑧= 4

4

𝑥−

6

𝑦+

5

𝑧= 1

6

𝑥+

9

𝑦−

20

𝑧= 2


𝐴 = [2 3 104 −6 56 9 −20

] , 𝑋 = [

1 𝑥⁄

1 𝑦⁄

1 𝑧⁄] and 𝐵 = [

412]

Now, |𝐴| = 2(120 − 45) − 3(−80 − 30) + 10(36 + 36) = 150 + 330 + 720 = 1200 ≠ 0

⇒ 𝐴−1 exists.

Therefore

𝐴11 = 75 𝐴12 = 110 𝐴13 = 72𝐴21 = 150 𝐴22 = −100 𝐴23 = 0𝐴31 = 75 𝐴32 = 30 𝐴33 = −24



1

1200[75 150 75110 −100 3072 0 −24

]

Also, 𝑋 = 𝐴−1𝐵

⇒ [

1 𝑥⁄

1 𝑦⁄

1 𝑧⁄] =

1

1200[75 150 75110 −100 3072 0 −24

] [412]

⇒

[ 1

𝑥1

𝑦1

𝑧]

=1

1200[300 + 150 + 150440 − 100 + 60288 + 0 − 48

]



⇒

[ 1

𝑥1

𝑦

1

𝑧]

=1

1200[600400240

] =

[ 1

21

31

5]

⇒1

𝑥=

1

2,

1

𝑦=

1

3,

1

𝑧=

1

5⇒ 𝑥 = 2, 𝑦 = 3, 𝑧 = 5

17. Choose the correct answer.

If 𝑎, 𝑏, 𝑐 are in A.P., then the determinant |𝑥 + 2 𝑥 + 3 𝑥 + 2𝑎𝑥 + 3 𝑥 + 4 𝑥 + 2𝑏𝑥 + 4 𝑥 + 5 𝑥 + 2𝑐

| is:

(A) 0

(B) 1

(C) 𝑥

(D) 2𝑥

Solution:

Given that 𝑎, 𝑏, 𝑐 are in A.P and determinant is |𝑥 + 2 𝑥 + 3 𝑥 + 2𝑎𝑥 + 3 𝑥 + 4 𝑥 + 2𝑏𝑥 + 4 𝑥 + 5 𝑥 + 2𝑐

|

= |𝑥 + 2 𝑥 + 3 𝑥 + 2𝑎

0 0 2(2𝑏 − 𝑎 − 𝑐)𝑥 + 4 𝑥 + 5 𝑥 + 2𝑐

| [Applying 𝑅2 → 2𝑅2 − (𝑅1 − 𝑅3)]

= |𝑥 + 2 𝑥 + 3 𝑥 + 2𝑎

0 0 0𝑥 + 4 𝑥 + 5 𝑥 + 2𝑐

| [∵ 𝑎, 𝑏, 𝑐 are in AP, therefore 2𝑏 = 𝑎 + 𝑐]

= 0 [∵ All the elements of 𝑅2 is zero]

Hence, the option (𝐴) is correct.


If 𝑥, 𝑦, 𝑧 are nonzero real numbers, then the inverse of matrix 𝐴 = [𝑥 0 00 𝑦 00 0 𝑧

] is:

(A) [𝑥−1 0 00 𝑦−1 0

0 0 𝑧−1

]



(B) 𝑥𝑦𝑧 [𝑥−1 0 00 𝑦−1 0

0 0 𝑧−1

]

(C) 1

𝑥𝑦𝑧[𝑥 0 00 𝑦 00 0 𝑧

]

(D) 1

𝑥𝑦𝑧[1 0 00 1 00 0 1

]

Solution:

Given matrix is 𝐴 = [𝑥 0 00 𝑦 00 0 𝑧

]

Now, |𝐴| = 𝑥(𝑦𝑧 − 0) − 0(0 − 0) + 0(0 − 0) = 𝑥𝑦𝑧 ≠ 0 ⇒ 𝐴−1 exists. Therefore,

𝐴11 = 𝑦𝑧 𝐴12 = 0 𝐴13 = 0𝐴21 = 0 𝐴22 = 𝑥𝑧 𝐴23 = 0𝐴31 = 0 𝐴32 = 0 𝐴33 = 𝑥𝑦



=1

𝑥𝑦𝑧[

𝑦𝑧 0 00 𝑥𝑧 00 0 𝑥𝑦

]

=

[ 1

𝑥0 0

01

𝑦0

0 01

𝑧]

= [𝑥−1 0 00 𝑦−1 0

0 0 𝑧−1

]

Hence, the option (𝐴) is correct.


Let 𝐴 = [1 sin 𝜃 1

−sin 𝜃 1 sin 𝜃−1 − sin 𝜃 1

], where 0 ≤ 𝜃 ≤ 2𝜋 then:

(A) det(𝐴) = 0

(B) det(𝐴) ∈ (2,∞)

(C) det(𝐴) ∈ (2, 4)

(D) det(𝐴) ∈ [2, 4]



Solution:

Given matrix is 𝐴 = [1 sin 𝜃 1

−sin 𝜃 1 sin 𝜃−1 −sin 𝜃 1

]

= 1(1 + sin2 𝜃) + sin𝜃(− sin 𝜃 + sin𝜃) + 1(sin2 𝜃 + 1) [Expanding along 𝐶1]

= 2(1 + sin2 𝜃)

Now, given that: 0 ≤ 𝜃 ≤ 2𝜋

⇒ 0 ≤ sin𝜃 ≤ 1

⇒ 0 ≤ sin2 𝜃 ≤ 1

⇒ 1 ≤ 1 + sin2 𝜃 ≤ 2

⇒ 2 ≤ 2(1 + sin2 𝜃) ≤ 4

⇒ det(𝐴) ∈ [2, 4]


CBSE NCERT Solutions for Class 12 Maths Chapter 04€¦ · Class-XII-Maths Determinants 1 Practice...

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Transcript of CBSE NCERT Solutions for Class 12 Maths Chapter 04€¦ · Class-XII-Maths Determinants 1 Practice...