Tag: Chapter 4 NCERT Maths Class 12th Solution

Exercise-4.4, Class 12th, Maths, Part -2

Question 14 :

For the matrix $A = [\begin{array}{cc} 3 & 2 \\ 1 & 1 \end{array}]$ , find the numbers $a$ and $b$ such that

$A^{2} + a A + b I = 0.$

Solution:

Given:

$A = [\begin{array}{cc} 3 & 2 \\ 1 & 1 \end{array}]$

We are to find $a$ and $b$ such that

$A^{2} + a A + b I = 0$

Step 1: Compute $A^{2}$

$A^{2} = [\begin{array}{cc} 3 & 2 \\ 1 & 1 \end{array}] [\begin{array}{cc} 3 & 2 \\ 1 & 1 \end{array}]$ Perform the multiplication:

$A^{2} = [\begin{array}{cc} 3 \times 3 + 2 \times 1 & 3 \times 2 + 2 \times 1 \\ 1 \times 3 + 1 \times 1 & 1 \times 2 + 1 \times 1 \end{array}] = [\begin{array}{cc} 9 + 2 & 6 + 2 \\ 3 + 1 & 2 + 1 \end{array}] = [\begin{array}{cc} 11 & 8 \\ 4 & 3 \end{array}]$

Step 2: Substitute in the equation

$A^{2} + a A + b I = 0$

That means:

$[\begin{array}{cc} 11 & 8 \\ 4 & 3 \end{array}] + a [\begin{array}{cc} 3 & 2 \\ 1 & 1 \end{array}] + b [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}]$

Step 3: Combine all matrices

$[\begin{array}{cc} 11 + 3 a + b & 8 + 2 a \\ 4 + a & 3 + a + b \end{array}] = [\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}]$

Step 4: Equate corresponding elements to zero

From the first row, first column:

$11 + 3 a + b = 0 (1)$

From the first row, second column:

$8 + 2 a = 0 (2)$

From the second row, first column:

$4 + a = 0 (3)$

From the second row, second column:

$3 + a + b = 0 (4)$

Step 5: Solve for $a$ and $b$

From (3):

$a = - 4$

Substitute $a = - 4$ :

$3 + (- 4) + b = 0 ⟹ b = 1$

So $a = - 4$ and $b = 1$ .

Step 6: Verification

Check equation (1):

$11 + 3 (- 4) + 1 = 11 - 12 + 1 = 0$

True. Check (2):

$8 + 2 (- 4) = 8 - 8 = 0$ True.

Final Answer:

$a = - 4, b = 1$

Question 15

For the matrix $A = [\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{array}]$ , show that

$A^{3} - 6 A^{2} + 5 A + 11 I = O .$

Hence, find $A^{- 1}$ .

Solution

Step 1 — compute $A^{2}$ and $A^{3}$ .

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

Compute $A^{2} = A \cdot A$ :

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

Compute $A^{3} = A^{2} \cdot A$ :

$A^{3} = [\begin{array}{ccc} 8 & 7 & 1 \\ - 23 & 27 & - 69 \\ 32 & - 13 & 58 \end{array}]$

Step 2 — form the combination $A^{3} - 6 A^{2} + 5 A + 11 I$ .

Calculate termwise:

$\begin{aligned} A^{3} - 6 A^{2} + 5 A + 11 I & = [\begin{array}{ccc} 8 & 7 & 1 \\ - 23 & 27 & - 69 \\ 32 & - 13 & 58 \end{array}] - 6 [\begin{array}{ccc} 4 & 2 & 1 \\ - 3 & 8 & - 14 \\ 7 & - 3 & 14 \end{array}] + 5 [\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{array}] + 11 [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ = [\begin{array}{ccc} 8 - 24 + 5 + 11 & 7 - 12 + 5 + 0 & 1 - 6 + 5 + 0 \\ - 23 + 18 + 5 + 0 & 27 - 48 + 10 + 11 & - 69 + 84 - 15 + 0 \\ 32 - 42 + 10 + 0 & - 13 + 18 - 5 + 0 & 58 - 84 + 15 + 11 \end{array}] \\ = [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] \end{aligned}$

So the relation is verified:

$A^{3} - 6 A^{2} + 5 A + 11 I = O$

Step 3 — find $A^{- 1}$ .

Starting from

$A^{3} - 6 A^{2} + 5 A + 11 I = O,$

multiply on the right by $A^{- 1}$ (which is valid because $A$ is invertible if we can find $A^{- 1}$ from this equation):

$A^{2} - 6 A + 5 I + 11 A^{- 1} = O$

Rearrange to solve for $A^{- 1}$ :

$11 A^{- 1} = - A^{2} + 6 A - 5 I, A^{- 1} = \frac{1}{11} (6 A - A^{2} - 5 I)$

Now substitute the matrices $A$ and $A^{2}$ :

$6 A - A^{2} - 5 I = 6 [\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{array}] - [\begin{array}{ccc} 4 & 2 & 1 \\ - 3 & 8 & - 14 \\ 7 & - 3 & 14 \end{array}] - 5 [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ $= [\begin{array}{ccc} 6 - 4 - 5 & 6 - 2 - 0 & 6 - 1 - 0 \\ 6 - (- 3) - 0 & 12 - 8 - 5 & - 18 - (- 14) - 0 \\ 12 - 7 - 0 & - 6 - (- 3) - 0 & 18 - 14 - 5 \end{array}] = [\begin{array}{ccc} - 3 & 4 & 5 \\ 9 & - 1 & - 4 \\ 5 & - 3 & - 1 \end{array}]$

Thus

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

You can write it explicitly as

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

(One may verify $A A^{- 1} = I$ )

Question 16 : If

$A = [\begin{array}{ccc} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{array}],$

verify that

$A^{3} - 6 A^{2} + 9 A - 4 I = O,$

and hence find $A^{- 1}$ .

Solution

Step 1 — compute $A^{2}$ .

$A^{2} = A \cdot A = [\begin{array}{ccc} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{array}] [\begin{array}{ccc} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{array}] = [\begin{array}{ccc} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{array}]$ (Each entry calculated: e.g. top-left $= 2 \cdot 2 + (- 1) (- 1) + 1 \cdot 1 = 4 + 1 + 1 = 6$ , top-middle $= 2 (- 1) + (- 1) 2 + 1 (- 1) = - 2 - 2 - 1 = - 5$

Step 2 — compute $A^{3}$ .

$A^{3} = A^{2} \cdot A = [\begin{array}{ccc} 22 & - 21 & 21 \\ - 21 & 22 & - 21 \\ 21 & - 21 & 22 \end{array}] .$

(For example, top-left $= 6 \cdot 2 + (- 5) (- 1) + 5 \cdot 1 = 12 + 5 + 5 = 22$ , top-middle $= 6 (- 1) + (- 5) 2 + 5 (- 1) = - 6 - 10 - 5 = - 21$

Step 3 — form the combination $A^{3} - 6 A^{2} + 9 A - 4 I$

Compute termwise:

$\begin{aligned} A^{3} - 6 A^{2} + 9 A - 4 I & = [\begin{array}{ccc} 22 & - 21 & 21 \\ - 21 & 22 & - 21 \\ 21 & - 21 & 22 \end{array}] - 6 [\begin{array}{ccc} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{array}] + 9 [\begin{array}{ccc} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{array}] - 4 [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ = [\begin{array}{ccc} 22 - 36 + 18 - 4 & - 21 + 30 - 9 + 0 & 21 - 30 + 9 + 0 \\ - 21 + 30 - 9 + 0 & 22 - 36 + 18 - 4 & - 21 + 30 - 9 + 0 \\ 21 - 30 + 9 + 0 & - 21 + 30 - 9 + 0 & 22 - 36 + 18 - 4 \end{array}] \\ = [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] \end{aligned}$

Thus the identity is verified:

$A^{3} - 6 A^{2} + 9 A - 4 I = O$

Step 4 — find $A^{- 1}$ .

Start from the matrix polynomial

$A^{3} - 6 A^{2} + 9 A - 4 I = O$

Right-multiply by $A^{- 1}$ (valid since the relation will imply invertibility):

$A^{2} - 6 A + 9 I - 4 A^{- 1} = O$

Rearrange to solve for $A^{- 1}$ :

$4 A^{- 1} = A^{2} - 6 A + 9 I \Rightarrow A^{- 1} = \frac{1}{4} (A^{2} - 6 A + 9 I)$

Now substitute the matrices $A$ and $A^{2}$ :

$A^{2} - 6 A + 9 I = [\begin{array}{ccc} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{array}] - 6 [\begin{array}{ccc} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{array}] + 9 [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ Compute it:

$A^{2} - 6 A + 9 I = [\begin{array}{ccc} 6 - 12 + 9 & - 5 + 6 + 0 & 5 - 6 + 0 \\ - 5 + 6 + 0 & 6 - 12 + 9 & - 5 + 6 + 0 \\ 5 - 6 + 0 & - 5 + 6 + 0 & 6 - 12 + 9 \end{array}] = [\begin{array}{ccc} 3 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3 \end{array}]$

Therefore

$A^{- 1} = \frac{1}{4} [\begin{array}{ccc} 3 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3 \end{array}] = [\begin{array}{ccc} \frac{3}{4} & \frac{1}{4} & - \frac{1}{4} \\ \frac{1}{4} & \frac{3}{4} & \frac{1}{4} \\ - \frac{1}{4} & \frac{1}{4} & \frac{3}{4} \end{array}]$

(You can verify $A A^{- 1} = I$ directly.)

Final answer:

$A^{3} - 6 A^{2} + 9 A - 4 I = O, A^{- 1} = [\begin{array}{ccc} \frac{3}{4} & \frac{1}{4} & - \frac{1}{4} \\ \frac{1}{4} & \frac{3}{4} & \frac{1}{4} \\ - \frac{1}{4} & \frac{1}{4} & \frac{3}{4} \end{array}]$

Question 17 :

Let $A$ be a nonsingular square matrix of order $3 \times 3$ . Then $∣ adj A ∣$ is equal to

(A) $∣ A ∣$
(B) $∣ A ∣^{2}$
(C) $∣ A ∣^{3}$
(D) $3 ∣ A ∣$

Solution:

We know the formula for the determinant of the adjugate (adjoint) of a square matrix:

$∣ adj A ∣ = ∣ A ∣^{n - 1}$ where $n$ is the order of the square matrix.

Here, $A$ is of order $3 \times 3$ .
So $n = 3$

$∣ adj A ∣ = ∣ A ∣^{3 - 1} = ∣ A ∣^{2}$

Final Answer:

$∣ adj A ∣ = ∣ A ∣^{2}$

Hence, the correct option is (B) $∣ A ∣^{2}$

Question 18:

If $A$ is an invertible matrix of order $2$ , then $\det (A^{- 1})$ is equal to

(A) $\det (A)$
(B) $\frac{1}{\det (A)}$
(C) $1$
(D) $0$

Solution:

We know the fundamental property of determinants:

$\det (A^{- 1}) = \frac{1}{\det (A)}$

provided $A$ is invertible (that is, $\det (A) \neq 0$ ).

Final Answer:

$\det (A^{- 1}) = \frac{1}{\det (A)}$

Hence, the correct option is (B) $\frac{1}{\det (A)}$ .

October 29, 2025
Exercise-4.3, Class 12th, Maths, Chapter 4, NCERT
1. Write minors and cofactors of the elements of the following determinants:

(i) $Δ = ∣ \begin{array}{cc} 2 & 4 \\ 0 & 3 \end{array} ∣$

For a $2 \times 2$ determinant the minor $M_{i j}$ of element $a_{i j}$ is the determinant left after deleting its row and column; for $2 \times 2$ deleting a row & column leaves a $1 \times 1$ number.

Elements and their minors:
- $a_{11} = 2 : M_{11} = 3$ . Cofactor $A_{11} = (- 1)^{1 + 1} M_{11} = + 3$
- $a_{12} = 4 : M_{12} = 0$ . Cofactor $A_{12} = (- 1)^{1 + 2} M_{12} = - 0 = 0$
- $a_{21} = 0 : M_{21} = 4$ . Cofactor $A_{21} = (- 1)^{2 + 1} M_{21} = - 4$
- $a_{22} = 3 : M_{22} = 2$ . Cofactor $A_{22} = (- 1)^{2 + 2} M_{22} = + 2$
(You can check: expansion along first row gives $Δ = 2 \cdot A_{11} + 4 \cdot A_{12} = 2 \cdot 3 + 4 \cdot 0 = 6$ , and direct determinant $2 \cdot 3 - 4 \cdot 0 = 6$ )

(ii) $Δ = ∣ \begin{array}{cc} a & c \\ b & d \end{array} ∣$

Minors (each $1 \times 1$ entry):
- $M_{11} = d, A_{11} = + d$
- $M_{12} = b, A_{12} = (- 1)^{1 + 2} b = - b$
- $M_{21} = c, A_{21} = (- 1)^{2 + 1} c = - c$
- $M_{22} = a, A_{22} = + a$
(So cofactors matrix is $(\begin{array}{cc} d & - b \\ - c & a \end{array})$ )

2. Write minors and cofactors for the following $3 \times 3$ determinants:

(i) $I_{3} = ∣ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} ∣$

We give minors $M_{i j}$ and cofactors $A_{i j} = (- 1)^{i + j} M_{i j}$

Because $I_{3}$ is diagonal, minors are determinants of the $2 \times 2$ submatrices:

Row 1:
- $M_{11} = ∣ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} ∣ = 1, A_{11} = + 1$
- $M_{12} = ∣ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} ∣ = 0, A_{12} = - 0 = 0$
- $M_{13} = ∣ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} ∣ = 0, A_{13} = + 0 = 0$
Row 2:
- $M_{21} = ∣ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} ∣ = 0, A_{21} = - 0 = 0$
- $M_{22} = ∣ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} ∣ = 1, A_{22} = + 1$
- $M_{23} = ∣ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} ∣ = 0, A_{23} = - 0 = 0$
Row 3:
- $M_{31} = ∣ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} ∣ = 0, A_{31} = + 0 = 0$
- $M_{32} = ∣ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} ∣ = 0, A_{32} = - 0 = 0$
- $M_{33} = ∣ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} ∣ = 1, A_{33} = + 1$
(So adj $I_{3}$ = transpose of cofactor matrix = identity again.)

(ii) $Δ = ∣ \begin{array}{ccc} 1 & 0 & 4 \\ 3 & 5 & 1 \\ 0 & 1 & 2 \end{array} ∣$

Compute minors (each is a $2 \times 2$ determinant) and cofactors:

Row 1:
- $M_{11} = ∣ \begin{array}{cc} 5 & 1 \\ 1 & 2 \end{array} ∣ = 5 \cdot 2 - 1 \cdot 1 = 10 - 1 = 9, A_{11} = + 9$
- $M_{12} = ∣ \begin{array}{cc} 3 & 1 \\ 0 & 2 \end{array} ∣ = 3 \cdot 2 - 1 \cdot 0 = 6, A_{12} = (- 1)^{1 + 2} 6 = - 6$
- $M_{13} = ∣ \begin{array}{cc} 3 & 5 \\ 0 & 1 \end{array} ∣ = 3 \cdot 1 - 5 \cdot 0 = 3, A_{13} = + 3$
Row 2:
- $M_{21} = ∣ \begin{array}{cc} 0 & 4 \\ 1 & 2 \end{array} ∣ = 0 \cdot 2 - 4 \cdot 1 = - 4, A_{21} = (- 1)^{2 + 1} (- 4) = + 4$
- $M_{22} = ∣ \begin{array}{cc} 1 & 4 \\ 0 & 2 \end{array} ∣ = 1 \cdot 2 - 4 \cdot 0 = 2, A_{22} = + 2$
- $M_{23} = ∣ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} ∣ = 1 \cdot 1 - 0 \cdot 0 = 1, A_{23} = (- 1)^{2 + 3} 1 = - 1$
Row 3:
- $M_{31} = ∣ \begin{array}{cc} 0 & 4 \\ 5 & 1 \end{array} ∣ = 0 \cdot 1 - 4 \cdot 5 = - 20, A_{31} = (- 1)^{3 + 1} (- 20) = - 20$
  (note sign: $(- 1)^{4} = + 1$ , so $A_{31} = + (- 20) = - 20$
- $M_{32} = ∣ \begin{array}{cc} 1 & 4 \\ 3 & 1 \end{array} ∣ = 1 \cdot 1 - 4 \cdot 3 = 1 - 12 = - 11, A_{32} = (- 1)^{3 + 2} (- 11) = + 11$
- $M_{33} = ∣ \begin{array}{cc} 1 & 0 \\ 3 & 5 \end{array} ∣ = 1 \cdot 5 - 0 \cdot 3 = 5, A_{33} = (- 1)^{3 + 3} 5 = + 5$
(You can verify determinant by expansion: $Δ = 1 \cdot A_{11} + 0 \cdot A_{12} + 4 \cdot A_{13} = 1 \cdot 9 + 0 + 4 \cdot 3 = 9 + 12 = 21$

Direct check yields same.)

3. Using cofactors of elements of the second row, evaluate

$Δ = ∣ \begin{array}{ccc} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array} ∣$

We will expand along the second row: $Δ = a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23}$

Compute the cofactors (minors first):
- For $a_{21} = 2$ : $M_{21} = ∣ \begin{array}{cc} 3 & 8 \\ 2 & 3 \end{array} ∣ = 3 \cdot 3 - 8 \cdot 2 = 9 - 16 = - 7$
  $A_{21} = (- 1)^{2 + 1} M_{21} = - (- 7) = + 7$
- For $a_{22} = 0$ : $M_{22} = ∣ \begin{array}{cc} 5 & 8 \\ 1 & 3 \end{array} ∣ = 5 \cdot 3 - 8 \cdot 1 = 15 - 8 = 7$
  $A_{22} = (- 1)^{2 + 2} M_{22} = + 7$
- For $a_{23} = 1$ : $M_{23} = ∣ \begin{array}{cc} 5 & 3 \\ 1 & 2 \end{array} ∣ = 5 \cdot 2 - 3 \cdot 1 = 10 - 3 = 7$
  $A_{23} = (- 1)^{2 + 3} M_{23} = - 7$
Now expansion:

$Δ = 2 \cdot A_{21} + 0 \cdot A_{22} + 1 \cdot A_{23} = 2 \cdot 7 + 0 + 1 \cdot (- 7) = 14 - 7 = 7$

(You may check by any other expansion; result is $7$ .)

4. Using cofactors of elements of the third column, evaluate

$Δ = ∣ \begin{array}{ccc} 1 & 1 & 1 \\ x & y & z \\ y & z & x \end{array} ∣$

We expand along third column: $Δ = a_{13} A_{13} + a_{23} A_{23} + a_{33} A_{33}$
where $a_{13} = 1, a_{23} = z, a_{33} = x$

Compute minors and cofactors:
- $A_{13} = (- 1)^{1 + 3} M_{13} = (- 1)^{4} M_{13} = + M_{13}$
  $M_{13} = ∣ \begin{array}{cc} x & y \\ y & z \end{array} ∣ = x z - y^{2}$
- $A_{23} = (- 1)^{2 + 3} M_{23} = (- 1)^{5} (- M_{23}) = - M_{23} $ $(Simpler:$ $A_{23} = - M_{23}$ )
  $M_{23} = ∣ \begin{array}{cc} 1 & 1 \\ y & z \end{array} ∣ = 1 \cdot z - 1 \cdot y = z - y$
  So $A_{23} = - (z - y) = y - z$
- $A_{33} = (- 1)^{3 + 3} M_{33} = + M_{33}$
  $M_{33} = ∣ \begin{array}{cc} 1 & 1 \\ x & y \end{array} ∣ = 1 \cdot y - 1 \cdot x = y - x$
  So $A_{33} = y - x$
Now expand:

$Δ = 1 \cdot (x z - y^{2}) + z \cdot (y - z) + x \cdot (y - x)$

Simplify term-by-term:

$Δ = x z - y^{2} + z y - z^{2} + x y - x^{2} .$

Group like terms:

$Δ = - x^{2} + (x z + z y + x y) - (y^{2} + z^{2})$

We can rewrite symmetric grouping if desired. But we can also notice a factorization — rearrange as

$Δ = - (x^{2} + y^{2} + z^{2}) + (x y + y z + z x) .$

Thus

$Δ = (x y + y z + z x) - (x^{2} + y^{2} + z^{2})$

(That is the simplest closed form. You may also write $Δ = - \frac{1}{2} [(x - y)^{2} + (y - z)^{2} + (z - x)^{2}]$ — indeed expanding that gives the same value. So $Δ \leq 0$ and equals zero exactly when $x = y = z$ .)

5. If $Δ = [a_{i j}]$ is the $3 \times 3$ determinant and $A_{i j}$ are cofactors of $a_{i j}$ , then which of the following equals $Δ$ ?

Options:
- (A) $a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33}$
- (B) $a_{11} A_{11} + a_{12} A_{21} + a_{13} A_{31}$
- (C) $a_{21} A_{11} + a_{22} A_{12} + a_{23} A_{13}$
- (D) $a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}$
Solution / reasoning.

Standard properties of cofactors / expansions:
- Determinant expansion along the first row gives
  $Δ = a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13}$
- Expansion along the first column gives
  $Δ = a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}$
Option (D) exactly matches the expansion along the first column, so it equals $Δ$ .
Options (A), (B), (C) are mixed-index combinations that do not in general equal $Δ$ (they give either other identities or zero). For example, the sum of elements of one row multiplied by cofactors of a different row equals $0$ .

Therefore the correct choice is

$(D) a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}$
October 28, 2025

Tag: Chapter 4 NCERT Maths Class 12th Solution

Exercise-4.4, Class 12th, Maths, Part -2

Question 14 :

Given:

Step 1: Compute $A^{2}$

Step 2: Substitute in the equation

Step 3: Combine all matrices

Step 4: Equate corresponding elements to zero

Step 5: Solve for $a$ and $b$

Step 6: Verification

Question 15

Solution

Question 16 : If

Solution

Question 17 :

Solution:

Question 18:

Solution:

Exercise-4.3, Class 12th, Maths, Chapter 4, NCERT

1. Write minors and cofactors of the elements of the following determinants:

(i) $Δ = ∣ \begin{array}{cc} 2 & 4 \\ 0 & 3 \end{array} ∣$

(ii) $Δ = ∣ \begin{array}{cc} a & c \\ b & d \end{array} ∣$

2. Write minors and cofactors for the following $3 \times 3$ determinants:

(i) $I_{3} = ∣ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} ∣$

(ii) $Δ = ∣ \begin{array}{ccc} 1 & 0 & 4 \\ 3 & 5 & 1 \\ 0 & 1 & 2 \end{array} ∣$

3. Using cofactors of elements of the second row, evaluate

4. Using cofactors of elements of the third column, evaluate

5. If $Δ = [a_{i j}]$ is the $3 \times 3$ determinant and $A_{i j}$ are cofactors of $a_{i j}$ , then which of the following equals $Δ$ ?

Tag: Chapter 4 NCERT Maths Class 12th Solution

Exercise-4.4, Class 12th, Maths, Part -2

Question 14 :

Given:

Step 1: Compute A2A2

Step 2: Substitute in the equation

Step 3: Combine all matrices

Step 4: Equate corresponding elements to zero

Step 5: Solve for aa and bb

Step 6: Verification

Question 15

Solution

Question 16 : If

Solution

Question 17 :

Solution:

Question 18:

Solution:

Exercise-4.3, Class 12th, Maths, Chapter 4, NCERT

1. Write minors and cofactors of the elements of the following determinants:

(i) Δ=∣2403∣

(ii) Δ=∣acbd∣

2. Write minors and cofactors for the following 3×3 determinants:

(i) I3=∣100010001∣

(ii) Δ=∣104351012∣

3. Using cofactors of elements of the second row, evaluate

4. Using cofactors of elements of the third column, evaluate

5. If Δ=[aij] is the 3×3 determinant and Aij​ are cofactors of aij, then which of the following equals Δ?

Step 1: Compute $A^{2}$

Step 5: Solve for $a$ and $b$

(i) $Δ = ∣ \begin{array}{cc} 2 & 4 \\ 0 & 3 \end{array} ∣$

(ii) $Δ = ∣ \begin{array}{cc} a & c \\ b & d \end{array} ∣$

2. Write minors and cofactors for the following $3 \times 3$ determinants:

(i) $I_{3} = ∣ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} ∣$

(ii) $Δ = ∣ \begin{array}{ccc} 1 & 0 & 4 \\ 3 & 5 & 1 \\ 0 & 1 & 2 \end{array} ∣$

5. If $Δ = [a_{i j}]$ is the $3 \times 3$ determinant and $A_{i j}$ are cofactors of $a_{i j}$ , then which of the following equals $Δ$ ?