Find inverse of 2×2 matrix [[3,1],[5,2]] and verify

Question

Find the inverse of the matrix $A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}$ and verify that $A \cdot A^{-1} = I$ .

Solution — Step by Step

For a 2×2 matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the determinant is $ad - bc$ .

\det(A) = (3)(2) - (1)(5) = 6 - 5 = 1

Since $\det(A) = 1 \neq 0$ , the inverse exists.

For a 2×2 matrix, the adjugate is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements:

\text{adj}(A) = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}

(swap 3 and 2, negate 1 and 5)

A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A) = \frac{1}{1} \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}

A \cdot A^{-1} = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}

Row 1 × Col 1: $(3)(2) + (1)(-5) = 6 - 5 = 1$

Row 1 × Col 2: $(3)(-1) + (1)(3) = -3 + 3 = 0$

Row 2 × Col 1: $(5)(2) + (2)(-5) = 10 - 10 = 0$

Row 2 × Col 2: $(5)(-1) + (2)(3) = -5 + 6 = 1$

A \cdot A^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \quad \checkmark

Why This Works

The inverse of a matrix “undoes” the transformation it represents. Just as $5 \times \frac{1}{5} = 1$ for numbers, $A \cdot A^{-1} = I$ for matrices.

The formula $A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A)$ comes from the property that $A \cdot \text{adj}(A) = \det(A) \cdot I$ . Dividing both sides by $\det(A)$ gives the inverse.

The determinant being 1 in this case made the computation especially clean — no fractions to deal with.

Alternative Method — Elementary Row Operations

We can find $A^{-1}$ by augmenting $A$ with the identity matrix and row-reducing:

\left[\begin{array}{cc|cc} 3 & 1 & 1 & 0 \\ 5 & 2 & 0 & 1 \end{array}\right]

$R_2 \to R_2 - \frac{5}{3}R_1$ :

\left[\begin{array}{cc|cc} 3 & 1 & 1 & 0 \\ 0 & \frac{1}{3} & -\frac{5}{3} & 1 \end{array}\right]

$R_2 \to 3R_2$ : $\left[\begin{array}{cc|cc} 3 & 1 & 1 & 0 \\ 0 & 1 & -5 & 3 \end{array}\right]$

$R_1 \to R_1 - R_2$ : $\left[\begin{array}{cc|cc} 3 & 0 & 6 & -3 \\ 0 & 1 & -5 & 3 \end{array}\right]$

$R_1 \to \frac{1}{3}R_1$ : $\left[\begin{array}{cc|cc} 1 & 0 & 2 & -1 \\ 0 & 1 & -5 & 3 \end{array}\right]$

The right half gives $A^{-1} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}$ . Same answer confirmed.

Common Mistake

Students often write the adjugate incorrectly for 2×2 matrices. The correct rule: swap the main diagonal elements (top-left and bottom-right), then negate the anti-diagonal elements (top-right and bottom-left).

For $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ : adjugate is $\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ . A common error is negating the wrong pair — swapping instead of negating the off-diagonal.

Question

Solution — Step by Step

Why This Works

Alternative Method — Elementary Row Operations

Common Mistake

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