Question

If $\begin{bmatrix} a & 2 & 3 \\ b & 5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 13 \\ 12 & 11 \end{bmatrix}$, then $(a, b)$ is}

• A. $(1, -2)$
• B. $(-1, -4)$
• C. $(1, 3)$
• D. $(1, -4)$

Hint:

Matrix multiplication: $(AB)_{ij} = \sum_k A_{ik}B_{kj}$. Solve for unknowns by equating specific entries of the product to the given matrix.

Answer 1

The Correct Option is D

To determine the values of \(a\) and \(b\) that satisfy the given matrix equation, we need to multiply the matrices and compare the result to the given solution matrix.

Given matrices:

\(\begin{bmatrix} a & 2 & 3 \\ b & 5 & -1 \end{bmatrix}\)

\(\begin{bmatrix} 1 & 2 \\ 3 & 4 \\ -1 & 1 \end{bmatrix}\)

The product of these matrices is:

Matrix multiplication is done by taking the dot product of rows of the first matrix with columns of the second matrix:

• The element in the first row, first column: \(a \times 1 + 2 \times 3 + 3 \times (-1)\)
• The element in the first row, second column: \(a \times 2 + 2 \times 4 + 3 \times 1\)
• The element in the second row, first column: \(b \times 1 + 5 \times 3 + (-1) \times (-1)\)
• The element in the second row, second column: \(b \times 2 + 5 \times 4 + (-1) \times 1\)

We know the product is equal to:

\(\begin{bmatrix} 4 & 13 \\ 12 & 11 \end{bmatrix}\)

Now, equating elements, we get:

1. \(a + 6 - 3 = 4\) → \(a + 3 = 4\) → \(a = 1\)
2. \(2a + 8 + 3 = 13\) → \(2a + 11 = 13\) → \(2a = 2\) → \(a = 1\) (again supporting \(a = 1\))
3. \(b + 15 + 1 = 12\) → \(b + 16 = 12\) → \(b = -4\)
4. \(2b + 20 - 1 = 11\) → \(2b + 19 = 11\) → \(2b = -8\) → \(b = -4\) (confirming \(b = -4\))

Hence, the correct solution for \((a, b)\) is \((1, -4)\).