Solveeit Logo
Login

Question

Mathematics Question on Matrices

If A=[23 1−4]A = \begin{bmatrix} 2 & 3 \\\ 1 & -4 \end{bmatrix} and B=[1−2 −13]B = \begin{bmatrix} 1 & -2 \\\ -1 & 3 \end{bmatrix}, then B−1A−1B^{-1} A^{-1} is equal to:

A

−111[145 51]-\frac{1}{11} \begin{bmatrix} 14 & 5 \\\ 5 & 1 \end{bmatrix}

B

111[1511 10]\frac{1}{11} \begin{bmatrix} 15 & 11 \\\ 1 & 0 \end{bmatrix}

C

111[145 51]\frac{1}{11} \begin{bmatrix} 14 & 5 \\\ 5 & 1 \end{bmatrix}

D

−111[1511 10]-\frac{1}{11} \begin{bmatrix} 15 & 11 \\\ 1 & 0 \end{bmatrix}

Answer

111[145 51]\frac{1}{11} \begin{bmatrix} 14 & 5 \\\ 5 & 1 \end{bmatrix}

Explanation

Solution

First, compute the inverses of A and B :

The determinant of A is:

det(A)=(2)(−4)−(3)(1)=−8−3=−11.\text{det}(A) = (2)(-4) - (3)(1) = -8 - 3 = -11.

The inverse of A is:

A−1=1−11[−4−3 −12]=111[43 1−2].A^{-1} = \frac{1}{-11} \begin{bmatrix} -4 & -3 \\\ -1 & 2 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 4 & 3 \\\ 1 & -2 \end{bmatrix}.

The determinant of B is:

det(B)=(1)(3)−(−2)(−1)=3−2=1.\text{det}(B) = (1)(3) - (-2)(-1) = 3 - 2 = 1.

The inverse of B is:

B−1=11[32 11]=[32 11].B^{-1} = \frac{1}{1} \begin{bmatrix} 3 & 2 \\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\\ 1 & 1 \end{bmatrix}.

Now compute B−1A−1B^{-1}A^{-1}:

B−1A−1=[32 11]⋅111[43 1−2].B^{-1}A^{-1} = \begin{bmatrix} 3 & 2 \\\ 1 & 1 \end{bmatrix} \cdot \frac{1}{11} \begin{bmatrix} 4 & 3 \\\ 1 & -2 \end{bmatrix}.

Perform the matrix multiplication:

[32 11]⋅[43 1−2]=[(3)(4)+(2)(1)(3)(3)+(2)(−2) (1)(4)+(1)(1)(1)(3)+(1)(−2)].\begin{bmatrix} 3 & 2 \\\ 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 4 & 3 \\\ 1 & -2 \end{bmatrix} = \begin{bmatrix} (3)(4) + (2)(1) & (3)(3) + (2)(-2) \\\ (1)(4) + (1)(1) & (1)(3) + (1)(-2) \end{bmatrix}.

=[12+29−4 4+13−2]=[145 51].= \begin{bmatrix} 12 + 2 & 9 - 4 \\\ 4 + 1 & 3 - 2 \end{bmatrix} = \begin{bmatrix} 14 & 5 \\\ 5 & 1 \end{bmatrix}.

Thus:

B−1A−1=111[145 51].B^{-1}A^{-1} = \frac{1}{11} \begin{bmatrix} 14 & 5 \\\ 5 & 1 \end{bmatrix}.