Important Questions For CBSE Class 12 Maths Chapter 4 Determinates

CBSE Class 12 Maths Chapter 4 Determinates Important Questions

Question 1. Three points P(2x, x + 3), Q(0, x) and R(x + 3, x + 6) are eollinear. then x is equal to

Solution: 4. 1

As per the given condition $\left|\begin{array}{ccc}
2 x & x+3 & 1 \\
0 & x & 1 \\
x+3 & x+6 & 1
\end{array}\right|=0$

[ar(ΔPQR) = 0]

⇒ 2x (x – x – 6) — 0(x + 3 – x — 6) + (x + 3) {x + 3 – x} =0

or -12x + 3x + 9 = 0 => -9x = -9 => x = 1

Read and Learn More CBSE Class 12 Maths Important Question and Answers

Question 2. If C_ij denotes the cofactor of element p_ij of the matrix P = $\left[\begin{array}{ccc}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & 2 & 4
\end{array}\right]$ , then the value of C₃₁, C₂₃ is:

5
24
-4
-5

Solution: 1. 5

P = $\left[\begin{array}{ccc}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & 2 & 4
\end{array}\right]$ (given)

⇒ C₃₁, C₂₃ = (3-4), {(-1)(2+3)} = 5

Determinants Class 12 Important Questions

Question 3. The system of linear equations 5x + ky = 5; 3x + 3y = 5 will be consistent if:

k ≠ -3
k = -5
k = 5
k ≠ 5

Solution: 4. k ≠ 5

The system of linear equations is given as:

5x + ky = 5 and 3x + 3y = 5

It can be written in matrix form as $\left[\begin{array}{ll}
5 & k \\
3 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
5 \\
5
\end{array}\right]$

or A X = B

So; the given system of linear equations is consistent if |A| ≠ 0

⇒ $\left|\begin{array}{ll}
5 & \mathrm{k} \\
3 & 3
\end{array}\right|$ 0

⇒15 – 3 k ≠ 0

⇒ k ≠ 5

Important Questions For CBSE Class 12 Maths Chapter 4

Question 4. If, for the matrix $A=\left[\begin{array}{cc}
\alpha & -2 \\
-2 & \alpha
\end{array}\right]$,$\left|\mathrm{A}^2\right|=125$; then the value of α is:

3
-3
1
1

Solution: 1. 3

Determinants Class 12 Important Questions

Given $\left|A^3\right|=125$

⇒ $(|A|)^3=125$

(because $\left|A^n\right|=|A|^n$)

⇒ $\left(\alpha^2-4\right)^3=125$

⇒ $\left(\alpha^2-4\right)=5$

⇒ $\alpha^2=9$ or $\alpha= \pm 3$

Question 5. Let matrix X = [x_ij] is given by X = $\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right]$, Then, the matrix Y = [m_ij]. where m_ij = Minor of X_ij is:

$\left[\begin{array}{ccc}
7 & -5 & -3 \\
19 & 1 & -11 \\
-11 & 1 & 7
\end{array}\right]$
$\left[\begin{array}{ccc}
7 & -19 & -11 \\
5 & -1 & -1 \\
3 & 11 & 7
\end{array}\right]$
$\left[\begin{array}{ccc}
7 & 19 & -11 \\
-3 & 11 & 7 \\
-5 & -1 & -1
\end{array}\right]$
$\left[\begin{array}{ccc}
7 & 19 & -11 \\
-1 & -1 & 1 \\
-3 & -11 & 7
\end{array}\right]$

Solution:

X = $\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right]$

Now, Y = [m_ij]

⇒ Y = $\left[\begin{array}{lll}
m_{11} & m_{12} & m_{13} \\
m_{21} & m_{22} & m_{23} \\
m_{31} & m_{32} & m_{33}
\end{array}\right] \Rightarrow Y=\left[\begin{array}{ccc}
7 & 19 & -11 \\
-1 & -1 & 1 \\
-3 & -11 & 7
\end{array}\right]$

Determinants Class 12 Important Questions

Question 6. If x =-4 is a root of $\left|\begin{array}{lll}
x & 2 & 3 \\
1 & x & 1 \\
3 & 2 & x
\end{array}\right|$ = 0, then the sum of the other two roots is:

4
-3
2
5

Solution: 1. 4

⇒ $\left|\begin{array}{lll}
x & 2 & 3 \\
1 & x & 1 \\
3 & 2 & x
\end{array}\right|$ = 0 (given)

⇒ $x\left(x^2-2\right)-2(x-3)+3(2-3 x)=0$

⇒ $x^3-2 x-2 x+6+6-9 x=0$

⇒ $x^3-13 x+12=0$

⇒ $(x+4)\left(x^2-4 x+3\right)=0$

(because x=-4 is a root)

⇒ $(x+4)(x-1)(x-3)=0$

Hence; the sum of other two roots = 1 + 3 = 4

Determinants Class 12 Important Questions

Question 7. The inverse of the matrix $X=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]$

$24\left[\begin{array}{ccc}1 / 2 & 0 & 0 \\ 0 & 1 / 3 & 0 \\ 0 & 0 & 1 / 4\end{array}\right]$
$\frac{1}{24}\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\frac{1}{24}\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]$
$\left[\begin{array}{ccc}1 / 2 & 0 & 0 \\ 0 & 1 / 3 & 0 \\ 0 & 0 & 1 / 4\end{array}\right]$

Solution:

Given, $X=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]$

|X| = $2(12)-0+0=24 \neq 0$

⇒ $X^{-1}$ exists

Now, adj X = $\left[\begin{array}{ccc}12 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 6\end{array}\right]$

⇒ $X^{-1}=\frac{1}{|X|} \cdot \mathrm{adj} X$

⇒ $X^{-1}=\frac{1}{24}\left[\begin{array}{ccc}
12 & 0 & 0 \\
0 & 8 & 0 \\
0 & 0 & 6
\end{array}\right]=\left[\begin{array}{ccc}
1 / 2 & 0 & 0 \\
0 & 1 / 3 & 0 \\
0 & 0 & 1 / 4
\end{array}\right]$

Question 8. If A is square matrix of order 3 such that A(adj A) = $\left[\begin{array}{ccc}-2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2\end{array}\right]$, then find |A|.

Solution:

Given, A (adj A) = $\left[\begin{array}{ccc}-2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2\end{array}\right]$

We know that A(adj A)= |A| I

Now, A(adj A)= $-2\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] = -2I = |A| I$

⇒ |A|=-2

Important Question for Class 12 Maths Chapter 4

Question 9. If A is a non-singular square matrix of order 3 such that A² = 3A, then the value of |A| is?

-3
3
9
27

Solution:

Given, A² = 3A, |A| ≠ 0, order of A is 3

∴ |A²| = |3A|

⇒ |A|²= 3³|A|

(|A²| = |A|² and |KA| = Kⁿ|A|)

or |A| = 27

Question 10. If A is a square matrix satisfying A’A = I, write the value of |A|.

Solution:

A’A = I (given)

⇒ $\left|A^{\prime} A\right|=|I|$

⇒ $\left|A^{\prime}\right||A|=|I| \Rightarrow|A|^2=1$ (because $\left|A^{\prime}\right|=|A|$)

⇒ $|A|=1 \text { or }|A|=-1$

Question 11. Given, $A=\left[\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right]$, compute A^-1 and show that 2A^-1 = 9I – A.

Solution:

Given, $A=\left[\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right]$

|A| = $14-(12)=2 \neq 0$

Hence, A is invertible.

adj A = $\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right]$

Hence, $A^{-1}=\frac{1}{|A|} \cdot \mathrm{adj} \cdot(\mathrm{A})$

⇒ $\mathrm{A}^{-1}=\frac{1}{2}\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right] \text { or } 2 \mathrm{~A}^{-1}=\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right]$….(1)

Now, R.H.S. = $9 \mathrm{I}-\mathrm{A}$

⇒ $9\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right]=\left[\begin{array}{cc}
9 & 0 \\
0 & 9
\end{array}\right]-\left[\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right]=\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right]=2 A^{-1}$= L.H.S. [from (1)]

Hence proved.

Question 12. If $A=\left[\begin{array}{ccc}
3 & 4 & 2 \\
0 & 2 & -3 \\
1 & -2 & 6
\end{array}\right]$; find A^-1.

Important Question for Class 12 Maths Chapter 4

Hence solve the following system of equations: 3x + 4y + 2z = 8; 0x + 2y + 3z = 3 and x + 2y + 6z = -2 or,

If $A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]$ find $(A B)^{-1} \text {. }$

Solution:

Given, $A=\left[\begin{array}{ccc}
3 & 4 & 2 \\
0 & 2 & -3 \\
1 & -2 & 6
\end{array}\right]$

⇒ |A| = 3(12 -6)-4(0 + 3) + 2(0-2)= 18- 12-4=2*0

Hence, A^-1 exists.

Now, co-factors are given as:

C₁₁ = 6, C₁₂ = —3, C₁₃ = -2,

C₂₁=-28, C₂₂ = 16, C₂₃= 10,

C₃₁= -16, C₃₂ = 9, C₃₃ = 6

Hence adj A = $\left[\begin{array}{ccc}6 & -3 & -2 \\ -28 & 16 & 10 \\ -16 & 9 & 6\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc}6 & -28 & -16 \\ -3 & 16 & 9 \\ -2 & 10 & 6\end{array}\right]$

⇒ $A^{-1}=\frac{1}{|A|}(\mathrm{adj} A)=\frac{1}{2}\left[\begin{array}{ccc}
6 & -28 & -16 \\
-3 & 16 & 9 \\
-2 & 10 & 6
\end{array}\right]$….(1)

The system of linear equations is given as:

3x + 4y + 2z = 8

0x + 2y —3 = 3

x – 2y + 6z = -2

This system is written in matrix form as

⇒ $\left[\begin{array}{ccc}
3 & 4 & 2 \\
0 & 2 & -3 \\
1 & -2 & 6
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
8 \\
3 \\
-2
\end{array}\right]$

⇒ AX = B

⇒ X = $A^{-1} \cdot B$

⇒ X = $\frac{1}{2}\left[\begin{array}{ccc}
6 & -28 & -16 \\
-3 & 16 & 9 \\
-2 & 10 & 6
\end{array}\right] \cdot\left[\begin{array}{l}
8 \\
3 \\
-2
\end{array}\right]=\frac{1}{2}\left[\begin{array}{l}
48-84+32 \\
-24+48-18 \\
-16+30-12
\end{array}\right]$

[from (1) and (2)]

⇒ $\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{2}\left[\begin{array}{r}
-4 \\
6 \\
2
\end{array}\right]$

x=-2, y=3, z=1

Given, $A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$…(1)

and $B=\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]$

|B| = 1 (3 – 0) – 2(- 1 – 0) – 2(2 – 0) = 3 + 2 – 4 = 1 ≠ 0

Now, the co-factors of matrix B are given as

⇒ $C_{11}=3, C_{12}=1, C_{13}=2$

⇒ $C_{21}=2, C_{22}=1, C_{23}=2,$

⇒ $C_{31}=6, C_{32}=2, C_{33}=5$

∴ (adj)(B) = $\left[\begin{array}{lll}
3 & 1 & 2 \\
2 & 1 & 2 \\
6 & 2 & 5
\end{array}\right]^1=\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right]$

⇒ $B^{-1}=\frac{1}{|B|}(\mathrm{adj} B)=\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right]$….(2)

∴ (AB)^-1 = B^-1 A^-1

= $\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right] \cdot\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]$

=$\left[\begin{array}{ccc}
9-30+30 & -3+12-12 & 3-10+12 \\
3-15+10 & -1+6-4 & 1-5+4 \\
6-30+25 & -2+12-10 & 2-10+10
\end{array}\right]$

⇒ $(A B)^{-1}=\left[\begin{array}{ccc}
9 & -3 & 5 \\
-2 & 1 & 0 \\
1 & 0 & 2
\end{array}\right]$

(from (1) and (2))

⇒ $(\mathrm{AB})^{-1}=\left[\begin{array}{ccc}
9 & -3 & 5 \\
-2 & 1 & 0 \\
1 & 0 & 2
\end{array}\right]$

Question 13. If $A=\left[\begin{array}{ccc}5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6\end{array}\right]$, find A^-1 and use it to solve the following system of equations.

5x – y + 4z = 5

2x + 3y + 5z = 2

5x – 2y + 6z = -1

Solution:

Given, $A=\left[\begin{array}{ccc}5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6\end{array}\right]$

∴ |A| = $\left|\begin{array}{ccc}
5 & -1 & 4 \\
2 & 3 & 5 \\
5 & -2 & 6
\end{array}\right|$ = 5(18+10) + (12 – 25) + 4 (-4 – 15)= 140 – 13 – 76 = 51 ≠ 0.

Hence: A^-1 exists

Now co-factors of elements of A are:

⇒ $A_{11}=28, A_{12}=13, A_{13}=-19$

⇒ $A_{21}=-2, A_{22}=10, A_{23}=5$

⇒ $A_{31}=-17, A_{32}=-17, A_{33}=17$

∴ adj A = $\left[\begin{array}{ccc}
28 & 13 & -19 \\
-2 & 10 & 5 \\
-17 & -17 & 17
\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc}
28 & -2 & -17 \\
13 & 10 & -17 \\
-19 & 5 & 17
\end{array}\right]$

⇒ $A^{-1}=\frac{1}{|A|}(\mathrm{adj} A)=\frac{1}{51}\left[\begin{array}{ccc}
28 & -2 & -17 \\
13 & 10 & -17 \\
-19 & 5 & 17
\end{array}\right]$

Given system of equations are

5x – y + 4z = 5

2x + 3y + 5z = 2

5x – 2y + 6z = -1

This system is written in matrix form as

⇒ ${\left[\begin{array}{ccc}
5 & -1 & 4 \\
2 & 3 & 5 \\
5 & -2 & 6
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
5 \\
2 \\
-1
\end{array}\right] }$

⇒ AX = B ⇒ X=$A^{-1}$ B

⇒ ${\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{51}\left[\begin{array}{ccc}
28 & -2 & -17 \\
13 & 10 & -17 \\
-19 & 5 & 17
\end{array}\right]\left[\begin{array}{r}
5 \\
2 \\
-1
\end{array}\right]}$

or $\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{51}\left[\begin{array}{l}
140-4+17 \\
65+20+17 \\
-95+10-17
\end{array}\right]$

⇒ $\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{51}\left[\begin{array}{r}
153 \\
102 \\
-102
\end{array}\right]=\left[\begin{array}{r}
3 \\
2 \\
-2
\end{array}\right]$

⇒ x= 3, y = 2, z = -2

Question 14. Show that, for matrix A = $\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]$, A³-6 A²+5 A+11 I=O, Hence, find $A^{-1}$.

Or,

Using the matrix method, solve the following system of equations:

3x – 2y + 3z = 8

2x + y – z = 1

4x – 3y + 2z = 4

Important Question for Class 12 Maths Chapter 4

Solution:

Given A = $\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]$

To prove: A³ – 6A² + 5A + 11 I = O

⇒ $A^2=A \cdot A=\left[\begin{array}{rrr}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right] \cdot\left[\begin{array}{rrr}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]=\left[\begin{array}{ccc}
1+1+2 & 1+2-1 & 1-3+3 \\
1+2-6 & 1+4+3 & 1-6-9 \\
2-1+6 & 2-2-3 & 2+3+9
\end{array}\right]$

⇒ $A^2=\left[\begin{array}{rrr}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right]$

and $A^3= A^2 \cdot A=\left[\begin{array}{rrr}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right] \cdot\left[\begin{array}{rrr}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]$

–$\left[\begin{array}{rrr}
4+2+2 & 4+4-1 & 4-6+3 \\
-3+8-28 & -3+16+14 & -3-24-42 \\
7-3+28 & 7-6-14 & 7+9+42
\end{array}\right]$

⇒ $A³=\left[\begin{array}{ccc}
8 & 7 & 1 \\
-23 & 27 & -69 \\
32 & -13 & 58
\end{array}\right]$

L.H.S. = $A^3-6 A^2+5 A+11I$

= $\left[\begin{array}{ccc}
8 & 7 & 1 \\
-23 & 27 & -69 \\
32 & -13 & 58
\end{array}\right]-6\left[\begin{array}{ccc}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right]+5\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]+11\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]$

= $\left[\begin{array}{ccc}
8-24+5+11 & 7-12+5+0 & 1-6+5+0 \\
-23+18+5+0 & 27-48+10+11 & -69+84-15+0 \\
32-42+10+0 & -13+18-5+0 & 58-84+15+11
\end{array}\right]$

= $\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]=\mathrm{O}$ Zero matrix

Now, A³-6A²+5A+11 I = O

A³A^-1– 6A²A ^-1+ 5AA^-1 + 11 I A^-1 = OA^-1 (Post multiplying both sides by $A^{-1}$

A²(AA^-1)-6A(AA^-1)+5(AA^-1)+11 I A^-1 = OA^-1

A²-6A +5I + 11A^-1 = O

because AA^-1=I and OA^-1=O

⇒ $A^{-1}=\frac{-1}{11} \cdot\left(A^2-6 A+5I\right)=\frac{-1}{11}\left\{\left[\begin{array}{ccc}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right]-6\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]+5\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\right\}$

⇒ $A^{-1}=\frac{-1}{11}\left[\begin{array}{ccc}
4-6+5 & 2-6+0 & 1-6+0 \\
-3-6+0 & 8-12+5 & -14+18+0 \\
7-12+0 & -3+6+0 & 14-18+5
\end{array}\right]$

⇒ $A^{-1}=\left[\begin{array}{ccc}
-3 / 11 & 4 / 11 & 5 / 11 \\
9 / 11 & -1 / 11 & -4 / 11 \\
5 / 11 & -3 / 11 & -1 / 11
\end{array}\right]$

The given system of equations are:

3x -2y + 3 z = 8

2x + y – z = 1 and

4x – 3y + 2z = 4

Important Question for Class 12 Maths Chapter 4

By using the matrix method; the given system of equations can be written as; AX = B

where $A=\left[\begin{array}{ccc}3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}8 \\ 1 \\ 4\end{array}\right]$

Now; $|A|=\left|\begin{array}{ccc}3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2\end{array}\right|$=-3+16-30=-17 ≠0

Hence; $\mathrm{A}^{-1}$ exists.

Now; $A=\left[\begin{array}{ccc}3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2\end{array}\right]$

Co-factors are given as: $C_{11}=-1, \quad C_{12}=-8, C_{13}=-10, C_{21}=-5, C_{22}=-6, C_{23}=1, C_{31}=-1, C_{32}=9, C_{33}=7$

Hence, (adj)(A)= $\left[C_{i j}\right]^{\mathrm{T}}$

adj(A)= $\left[\begin{array}{ccc}
-1 & -5 & -1 \\
-8 & -6 & 9 \\
-10 & 1 & 7
\end{array}\right]$

Now; AX = $B \Rightarrow X=A^{-1} \cdot B$ = $\frac{\mathrm{adj} A}{|A|} \cdot B$

⇒ $\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=-\frac{1}{17}\left[\begin{array}{ccc}
-1 & -5 & -1 \\
-8 & -6 & 9 \\
-10 & 1 & 7
\end{array}\right] \cdot\left[\begin{array}{l}
8 \\
1 \\
4
\end{array}\right]=-\frac{1}{17}\left[\begin{array}{c}
-8-5-4 \\
-64-6+36 \\
-80+1+28
\end{array}\right]=-\frac{1}{17}\left[\begin{array}{l}
-17 \\
-34 \\
-51
\end{array}\right]$

⇒ $\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]$

x=1, y=2, z=3

Question 15. If $A=\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]$, find A^-1. Hence, using A^-1, solve the system of equations:

2x – 3y + 5z = 11,

3x + 2y – 4z = -5,

x + y -2z = -3.

Solution:

Given, $A=\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]$

A is invertible if |A| ≠ 0

Now. |A| = 2(-4 + 4) + 3 (-6 + 4) + 5(3 – 2)= 0 – 6 + 5 =-1≠0

Co-factors are given as :

⇒ $C_{11}=0, C_{12}=2, C_{13}=1$

⇒ $C_{21}=-1, C_{22}=-9, C_{23}=-5$

⇒ $C_{31}=2, C_{32}=23, C_{33}=13$

⇒ (adj) $(\mathrm{A})=\left[\mathrm{C}_{1 \mathrm{ij}}\right]^{\mathrm{T}}$

or (adj) $(\mathrm{A})=\left[\begin{array}{ccc}
0 & 2 & 1 \\
-1 & -9 & -5 \\
2 & 23 & 13
\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc}
0 & -1 & 2 \\
2 & -9 & 23 \\
1 & -5 & 13
\end{array}\right]$

⇒ $\mathrm{A}^{-1}=\frac{\mathrm{adj}(\mathrm{A})}{|\mathrm{A}|}=\left[\begin{array}{ccc}
0 & 1 & -2 \\
-2 & 9 & -23 \\
-1 & 5 & -13
\end{array}\right]$

Now, given equations are 2x-3y + 5z = 11; 3x + 2y -4z = -5 and x + y – 2z = -3

⇒ ${\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{l}
11 \\
-5 \\
-3
\end{array}\right]}$

⇒ $\mathrm{AX}=\mathrm{B} \text { or } \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B} \\
\Rightarrow \mathrm{X}=\left[\begin{array}{ccc}
0 & 1 & -2 \\
-2 & 9 & -23 \\
-1 & 5 & -13
\end{array}\right]\left[\begin{array}{l}
11 \\
-5 \\
-3
\end{array}\right]$

or $\mathrm{X}=\left[\begin{array}{c}
0 \times 11+(1) \times(-5)+(-2) \times(-3) \\
(-2) \times 11+9 \times(-5)+(-23) \times(-3) \\
(-1) \times 11+5 \times(-5)+(-13) \times(-3)
\end{array}\right] \\
\Rightarrow \quad\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]$

⇒ $\mathrm{x}=1, \mathrm{y}=2, \mathrm{z}=3$

CBSE Class 12 Maths Chapter 4 Determinates Important Questions

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