Prove a Matrix Equation and Hence Find the Inverse

This lesson proves that the matrix $A = \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix}$ satisfies the equation $A^2 - 3A - 7I = 0$, and then uses that relation to find $A^{-1}$ without ever computing a determinant.

Squaring the matrix

We start by working out $A^2 = A \cdot A$, multiplying each row of the first matrix into each column of the second.

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

Entry by entry:

• Top left: $5\cdot 5 + 3\cdot(-1) = 25 - 3 = 22$
• Top right: $5\cdot 3 + 3\cdot(-2) = 15 - 6 = 9$
• Bottom left: $(-1)\cdot 5 + (-2)\cdot(-1) = -5 + 2 = -3$
• Bottom right: $(-1)\cdot 3 + (-2)\cdot(-2) = -3 + 4 = 1$

$A^2 = \begin{bmatrix} 22 & 9 \\ -3 & 1 \end{bmatrix}$

Proving $A^2 - 3A - 7I = 0$

Now we form the full expression. Writing out each piece:

$-3A = \begin{bmatrix} -15 & -9 \\ 3 & 6 \end{bmatrix}, \qquad -7I = \begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix}$

Adding the three matrices together, corresponding entry by corresponding entry:

$A^2 - 3A - 7I = \begin{bmatrix} 22 - 15 - 7 & 9 - 9 + 0 \\ -3 + 3 + 0 & 1 + 6 - 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

Every entry is zero, so the equation holds and the first part is proved.

Finding the inverse

We reuse the same relation. Moving the identity term across gives

$A^2 - 3A = 7I.$

Factor an $A$ out of the left side:

$A(A - 3I) = 7I.$

Pre-multiplying both sides by $A^{-1}$ and using $A^{-1}A = I$:

$A - 3I = 7A^{-1}, \qquad \text{so} \qquad A^{-1} = \tfrac{1}{7}(A - 3I).$

Substituting the values

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

Therefore

$A^{-1} = \tfrac{1}{7}\begin{bmatrix} 2 & 3 \\ -1 & -5 \end{bmatrix}.$

As a check, $A \cdot A^{-1} = \tfrac{1}{7}\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = I$, confirming the answer.

Key takeaways

• Squaring a $2\times 2$ matrix is just the row-into-column rule applied to the matrix and itself.
• If a matrix satisfies $A^2 - 3A - 7I = 0$, it must be invertible, and the inverse comes straight from rearranging that equation.
• Factoring $A(A - 3I) = 7I$ and pre-multiplying by $A^{-1}$ gives $A^{-1} = \tfrac{1}{7}(A - 3I)$, no determinant needed.