(a) If $A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$, find $A^{-1}$ and hence solve the following system of linear equations : $2x - 3y + 5z = 11$, $3x + 2y - 4z = -5$, $x + y - 2z = -3$ OR (b) Using properties of determinants, prove that $\Delta = \begin{vmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (c+a)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{vmatrix} = 2\,abc\,(a + b + c)^3$

> [!TLDR] > (a) $|A| = -1$, so $A^{-1} = -\mathrm{adj}(A)$; then $X = A^{-1}B$ gives $x=1, y=2, z=3$. > (b) Column operations $C_1\to C_1-C_3$, $C_2\to C_2-C_3$ pull out $(a+b+c)^2$; a further row operation and expansion yield $2abc(a+b+c)^3$. **Part (a) — Inverse and system** > [!FORMULA] > $A^{-1} = \dfrac{1}{|A|}\,\mathrm{adj}(A)$ (exists when $|A|\ne0$); the system $AX=B$ has solution $X = A^{-1}B$. With $A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$: 1. **Determinant:** $|A| = 2(2\cdot(-2) - (-4)\cdot1) - (-3)(3\cdot(-2) - (-4)\cdot1) + 5(3\cdot1 - 2\cdot1)$ $= 2(-4+4) + 3(-6+4) + 5(3-2) = 0 - 6 + 5 = -1 \ (\ne 0)$, so $A^{-1}$ exists. 2. **Adjoint** (transpose of the cofactor matrix): $$\mathrm{adj}(A) = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix}$$ 3. **Inverse:** $A^{-1} = \dfrac{1}{-1}\,\mathrm{adj}(A) = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}$. 4. **Solve** with $B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}$, $X = A^{-1}B$: - $x = 0(11) + 1(-5) + (-2)(-3) = -5 + 6 = 1$ - $y = -2(11) + 9(-5) + (-23)(-3) = -22 - 45 + 69 = 2$ …

(a) If A = 2 -3 5 \3 2 -4 \1 1 -2 , find A⁻¹ and hence — CBSE Class 12 Applied Mathematics 2025 PYQ

Question

A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}

(OR)

(b) Using properties of determinants, prove that

\Delta = \begin{vmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (c+a)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{vmatrix} = 2\,abc\,(a + b + c)^3

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Concept understanding — Minor of a Matrix Element

Minor of a Matrix Element – From Intuition to Precision

Imagine you have a large matrix — say a $3 \times 3$ grid of numbers. You pick one specific element, like the number in the second row and third column. Now ask: if I wanted to ignore that element and everything in its row and column, what smaller matrix remains? That smaller matrix is the key.

The minor of an element is simply the determinant of that leftover submatrix. It captures the "influence" of that element when you strip away its entire row and column.

The Intuition First

Think of a matrix as a system of equations. Each row is an equation, each column a variable. When you focus on one element, you are asking: what happens to the rest of the system if I remove the equation and the variable that this element belongs to? The minor tells you the "size" (determinant) of that reduced system.

For a $2 \times 2$ matrix, the minor of any element is just the single number diagonally opposite it — because removing one row and one column leaves a $1 \times 1$ matrix, whose determinant is that number itself.

For a $3 \times 3$ matrix, the minor of an element is the determinant of a $2 \times 2$ matrix formed by the four numbers that are not in the same row or column as the chosen element.

The Precise Statement

Let $A$ be an $n \times n$ square matrix. Let $a_{ij}$ be the element in the $i$ -th row and $j$ -th column.

M_{ij} = \det(\text{matrix obtained by deleting row } i \text{ and column } j \text{ from } A)

$M_{ij}$ is called the minor of the element $a_{ij}$ .

A Concrete Example

Take the matrix:

A = \begin{pmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{pmatrix}

Find the minor of the element $a_{23} = 8$ (row 2, column 3).

Delete row 2 and column 3. What remains?

\begin{pmatrix} 1 & 4 \\ 3 & 6 \end{pmatrix}

The minor is the determinant of this $2 \times 2$ matrix:

M_{23} = (1)(6) - (4)(3) = 6 - 12 = -6

So the minor of $8$ is $-6$ .

Tip

The minor is not the element itself — it is the determinant of the submatrix left after removing that element's row and column. For a $1 \times 1$ matrix, the minor of the single element is $1$ (the determinant of an empty matrix is defined as $1$ ), but that's a special case.

Why Minors Matter

Minors are the building blocks of cofactors, which in turn are used to compute determinants of large matrices (Laplace expansion) and to find inverses. Every time you expand a determinant along a row or column, you are summing products of elements and their minors (with appropriate signs). …

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$|A| = -1$ , so $A^{-1} = -\mathrm{adj}(A)$ ; then $X = A^{-1}B$ gives $x=1, y=2, z=3$ .
Column operations $C_1\to C_1-C_3$ , $C_2\to C_2-C_3$ pull out $(a+b+c)^2$ ; a further row operation and expansion yield $2abc(a+b+c)^3$ .

Part (a) — Inverse and system

A^{-1} = \dfrac{1}{|A|}\,\mathrm{adj}(A)

With $A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$ :

Determinant: $|A| = 2(2\cdot(-2) - (-4)\cdot1) - (-3)(3\cdot(-2) - (-4)\cdot1) + 5(3\cdot1 - 2\cdot1)$ $= 2(-4+4) + 3(-6+4) + 5(3-2) = 0 - 6 + 5 = -1 \ (\ne 0)$ , so $A^{-1}$ exists.
Adjoint (transpose of the cofactor matrix):

\mathrm{adj}(A) = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix}

Inverse: $A^{-1} = \dfrac{1}{-1}\,\mathrm{adj}(A) = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}$ .
Solve with $B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}$ $B = 11 - 5 - 3$ , $X = A^{-1}B$ $X = A^{- 1} B$ :
- $x = 0(11) + 1(-5) + (-2)(-3) = -5 + 6 = 1$
- $y = -2(11) + 9(-5) + (-23)(-3) = -22 - 45 + 69 = 2$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set 465/RQPS/41 markMCQ
$\Delta = \begin{vmatrix} 42 & 2 & 5 \\ 79 & 7 & 9 \\ 29 & 5 & 3 \end{vmatrix}$
›Reveal solutionSolution
Cofactor expansion along row 1 yields $0$ , so $\Delta=0$ .

$\begin{vmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}=a_1(b_2c_3-b_3c_2)-a_2(b_1c_3-b_3c_1)+a_3(b_1c_2-b_2c_1)$

$\Delta=\begin{vmatrix}42&2&5\\79&7&9\\29&5&3\end{vmatrix}$ .

$42(7\cdot3-9\cdot5)=42(21-45)=42(-24)=-1008$ . …
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CBSE 2024Set 465/S/RQPS/41 markMCQ
$a_{13}$
›Reveal solutionSolution
Deleting row 1 and column 3 gives $\begin{vmatrix}1&2\\2&1\end{vmatrix}$ , so A is true; R reverses the row/column indices and is false.

$M_{ij}$

Element $a_{13}$ sits in row 1, column 3 ( $i=1,\ j=3$ ); its value is 6.

Delete row 1 and column 3 of $\begin{bmatrix}0&2&6\\1&2&-1\\2&1&3\end{bmatrix}$ : the remaining rows are $(1,2)$ and $(2,1)$ , giving $\begin{vmatrix}1&2\\2&1\end{vmatrix}$ . This matches the Assertion, so A is true. …
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CBSE 2023Set 465/EF1GH/41 markMCQ
$\begin{vmatrix} 2 & 3 & 2 \\ x & x & x \\ 4 & 9 & 1 \end{vmatrix} + 3 = 0$
(a) $-1$
(b) $0$
(c) $1$
(d) $3$
›Reveal solutionSolution
Factor $x$ from row 2; the remaining determinant is $3$ , so $3x+3=0 \Rightarrow x=-1$ .

$3\times3$

$\begin{vmatrix}2&3&2\\x&x&x\\4&9&1\end{vmatrix}=x\begin{vmatrix}2&3&2\\1&1&1\\4&9&1\end{vmatrix}$ (taking $x$ from row 2). …
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