Q.If $A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$, find $A^{-1}$. Using $A^{-1}$ solve the system of equations $2x - 3y + 5z = 11$ $3x + 2y - 4z = -5$ $x + y - 2z = -3$

Concept: Inverse Matrix Method — For a system $AX = B$, if $A$ is invertible, $X = A^{-1}B$.

Step 1: Find $|A|$

$|A| = 2(2 \cdot -2 - (-4) \cdot 1) - (-3)(3 \cdot -2 - (-4) \cdot 1) + 5(3 \cdot 1 - 2 \cdot 1)$

$= 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2) = 0 + 3(-2) + 5(1) = -6 + 5 = -1 \neq 0$, so $A^{-1}$ exists.

Step 2: Find adjoint of $A$

Cofactor matrix:

$C_{11} = (2 \cdot -2 - (-4) \cdot 1) = 0$, $C_{12} = -(3 \cdot -2 - (-4) \cdot 1) = -(-6+4)=2$,

$C_{13} = (3 \cdot 1 - 2 \cdot 1) = 1$,

$C_{21} = -((-3) \cdot -2 - 5 \cdot 1) = -(6 - 5) = -1$,

$C_{22} = (2 \cdot -2 - 5 \cdot 1) = -4 - 5 = -9$,

$C_{23} = -(2 \cdot 1 - (-3) \cdot 1) = -(2 + 3) = -5$,

$C_{31} = ((-3) \cdot -4 - 5 \cdot 2) = 12 - 10 = 2$,

$C_{32} = -(2 \cdot -4 - 5 \cdot 3) = -(-8 - 15) = 23$,

$C_{33} = (2 \cdot 2 - (-3) \cdot 3) = 4 + 9 = 13$.

Adjoint = transpose of cofactor matrix:

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

Step 3: Compute $A^{-1}$

$A^{-1} = \frac{\text{adj}(A)}{|A|} = \frac{1}{-1} \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}$. …

1. Determinant. For $A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$, expanding along the first row:

2. Cofactors $C_{ij}=(-1)^{i+j}M_{ij}$:

3. Adjoint (transpose of the cofactor matrix):

4. Inverse:

$A^{-1} = \frac{1}{-1}\,\text{adj}(A) = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}.$ …

Method: Finding $A^{-1}$ First, Then Reusing It to Solve $AX=B$

Some problems ask for the inverse of a matrix AND the solution of a related system in the same question. Since the system's coefficient matrix is exactly $A$, you only need to compute $A^{-1}$ once and reuse it — never invert twice.

Steps

Step 1: Compute $\det(A)$

Expand along the row or column with the most zeros to minimise arithmetic. Confirm $\det(A) \neq 0$ before continuing — otherwise no inverse exists.

Step 2: Build the cofactor matrix, then transpose it to get $\operatorname{adj}(A)$

Work through all nine cofactors $C_{ij}=(-1)^{i+j}M_{ij}$ systematically (row by row), then transpose the resulting matrix.

Step 3: Form $A^{-1} = \dfrac{1}{\det(A)}\operatorname{adj}(A)$ …

Common Mistakes

Mistake 1: Re-deriving $A$ (or re-inverting) from the system instead of reusing the already-computed inverse

Why it's wrong: when a question gives $A$ (or asks you to find $A^{-1}$) and then a "related" system, the coefficient matrix of that system IS $A$ — recomputing it from scratch wastes time and risks a fresh arithmetic error. Correct approach: confirm the system's coefficients match $A$'s rows, then plug your already-computed $A^{-1}$ straight into $X = A^{-1}B$.

Mistake 2: Losing track of a sign while transposing the cofactor matrix …