Q.$\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$ Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.

Concept: Inverse of a matrix via elementary row operations (Gauss-Jordan method).

We augment the given matrix with the identity and row-reduce until the left side becomes $I$.

Step 1: Write the augmented matrix $[A \mid I]$.

Step 2: Eliminate below the first pivot.

$R_2 \to R_2 - 3R_1$, $R_3 \to R_3 - R_1$:

Step 3: Swap $R_2$ and $R_3$ to get a pivot in row 2, then eliminate.

$R_2 \leftrightarrow R_3$:

$R_3 \to R_3 - 3R_2$:

Step 4: Back-substitute to obtain $I$ on the left.

$R_3 \to -\frac{1}{11}R_3$:

$R_2 \to R_2 - R_3$, $R_1 \to R_1 - 2R_3$: …

For $A=\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$, use $A^{-1}=\dfrac{1}{\det A}\,\operatorname{adj}(A)$.

1. Determinant (expand along column 2, which has two zeros):

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

$C_{21}=-\begin{vmatrix} -1 & 2 \\ 0 & 3 \end{vmatrix}=3,\quad C_{22}=\begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix}=1,\quad C_{23}=-\begin{vmatrix} 1 & -1 \\ 1 & 0 \end{vmatrix}=-1$ …

Method: Finding the Inverse of a $3\times3$ Matrix Using the Adjoint

The standard adjoint-method procedure for a general square matrix whose inverse is required.

Steps

Step 1: Compute $|A|$ and confirm it is nonzero

Expand along the row/column with the most zeros. If $|A|\ne0$, the inverse exists; if $|A|=0$, stop — the matrix is singular and has no inverse.

Step 2: Compute all nine cofactors $C_{ij}=(-1)^{i+j}M_{ij}$

Delete the relevant row and column for each entry to form its $2\times2$ minor, then apply the sign.

Step 3: Form the adjoint as the transpose of the cofactor matrix

$\operatorname{adj}(A) = \begin{bmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{bmatrix}.$ …

Common Mistakes

Mistake 1: Forgetting the transpose step when forming the adjoint from the cofactor matrix

Why it's wrong: the adjoint is the transpose of the cofactor matrix, not the cofactor matrix itself — using the untransposed version silently swaps off-diagonal entries and produces a wrong inverse. Correct approach: explicitly write $(\operatorname{adj} A)_{ij} = C_{ji}$ (swap row/column indices) when assembling the adjoint from the computed cofactors.

Mistake 2: Making a sign error in one of the nine cofactors, especially when a minor itself contains a negative entry …