Question

If $A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & a \\ 2 & 4 & 7 \end{bmatrix}$ and $B = \begin{bmatrix} 13 & 2 & b \\ -3 & -1 & 2 \\ -2 & 0 & 1 \end{bmatrix}$ where matrix $B$ is the inverse of matrix $A$, then the values of $a$ and $b$ are

• A. $a = -5, b = 7$
• B. $a = 7, b = -5$
• C. $a = -7, b = 5$
• D. $a = 5, b = -7$

Hint:

When a matrix equation involves the identity condition $A \cdot A^{-1} = I$, you don't need to compute the full inverse or expand the entire matrix multiplication! Pick rows and columns strategically to solve for each unknown variable independently in a single step.

Answer 1

The Correct Option is D

Step 1: Use the inverse relation.
Since $B = A^{-1}$, the product $AB$ must equal the identity matrix $I$.
Step 2: Target an entry that isolates $b$.
The $(3,3)$ entry of $AB$ must equal $1$: row $3$ of $A$ times column $3$ of $B$ gives $(2)(b) + (4)(2) + (7)(1) = 1$.
Step 3: Solve for $b$.
$2b + 8 + 7 = 1 \Rightarrow 2b = -14 \Rightarrow b = -7$.
Step 4: Target an entry that isolates $a$.
The $(2,1)$ entry of $AB$ must equal $0$: row $2$ of $A$ times column $1$ of $B$ gives $(1)(13) + (1)(-3) + (a)(-2) = 0$.
Step 5: Solve for $a$.
$13 - 3 - 2a = 0 \Rightarrow 10 = 2a \Rightarrow a = 5$.
Step 6: Verify consistency.
Checking the $(1,3)$ entry with $b=-7$: $(1)(-7)+(2)(2)+(3)(1) = -7+4+3 = 0$, exactly as $I$ requires. So $a = 5$, $b = -7$.
\[ \boxed{a = 5,\ b = -7} \]