Question

Let A=begin{bmatrix}1 & 2&-1 1&0&31&1&-1 end{bmatrix} i) Find the determinant of A ii)Verify that the matrix begin{bmatrix}-frac{3}{4} & frac{1}{4}&frac{6}{4} 1&0&-1frac{1}{4}&frac{1}{4}&-frac{1}{2} end{bmatrix} is the inverse matrix of A

Matrices
ANSWERED
asked 2021-02-14
Let \(A=\begin{bmatrix}1 & 2&-1 \\1&0&3\\1&1&-1 \end{bmatrix}\)
i) Find the determinant of A
ii)Verify that the matrix
\(\begin{bmatrix}-\frac{3}{4} & \frac{1}{4}&\frac{6}{4} \\1&0&-1\\\frac{1}{4}&\frac{1}{4}&-\frac{1}{2} \end{bmatrix}\) is the inverse matrix of A

Answers (1)

2021-02-15
Step 1
For part (i) of a:
According to the given information, it is required to calculate the determinant of matrix
\(A=\begin{bmatrix}1 & 2&-1 \\1&0&3\\1&1&-1 \end{bmatrix}\)
For this use the formula
\(\begin{vmatrix}x_1 & y_1&z_1 \\x_2&y_2&z_2\\x_3&y_3&z_3\end{vmatrix}=x_1\begin{vmatrix}y_2 & z_2 \\y_3 & z_3 \end{vmatrix}-y_1\begin{vmatrix}x_2 & z_2 \\x_3 & z_3 \end{vmatrix}+z_1\begin{vmatrix}x_2 & y_2 \\x_3 & y_3 \end{vmatrix}\)
Step 2
Using above formula determinant of A can be calculated as
\(|A|=\begin{vmatrix}1 & 2&-1 \\1&0&3\\1&1&-1 \end{vmatrix}\)
\(=1\begin{vmatrix}0 & 3 \\1 & -1 \end{vmatrix}-2\begin{vmatrix}1 & 3 \\1 & -1 \end{vmatrix}-1\begin{vmatrix}1 & 0 \\1 & 1 \end{vmatrix}\)
\(=1(0\times-1-3\times1)-2(1\times-1-3\times1)-1(1\times1-0\times1)\)
=1(-3)-2(-1-3)-1
=-3+8-1
=4
Thus, determinant of matrix A is 4.
Step 3
For part (ii) of a:
To verify that the matrix
\(\begin{bmatrix}-\frac{3}{4} & \frac{1}{4}&\frac{6}{4} \\1&0&-1\\\frac{1}{4}&\frac{1}{4}&-\frac{1}{2} \end{bmatrix}\)
is inverse of given matrix A use the definition of identity matrix that states:
\(AA^{-1}=I\)
Step 4
Now calculate the multiplication of these two matrices, as
\(\begin{bmatrix}1 & 2&-1 \\1&0&3\\1&1&-1 \end{bmatrix}\begin{bmatrix}-\frac{3}{4} & \frac{1}{4}&\frac{6}{4} \\1&0&-1\\\frac{1}{4}&\frac{1}{4}&-\frac{1}{2} \end{bmatrix}=\begin{bmatrix}1\times-\frac{3}{4}+2\times1-1\times\frac{1}{4} & 1\times\frac{1}{4}+2\times0-1\times\frac{1}{4}&1\times\frac{6}{4}+2\times-1-1\times-\frac{1}{2} \\1\times-\frac{3}{4}+0\times1+3\times\frac{1}{4} & 1\times\frac{1}{4}+0\times0+3\times\frac{1}{4}&1\times\frac{6}{4}+0\times-1+3\times-\frac{1}{2}\\1\times-\frac{3}{4}+1\times1-1\times\frac{1}{4}&1\times\frac{1}{4}+1\times0-1\times\frac{1}{4}&1\times\frac{6}{4}+1\times-1-1\times-\frac{1}{2} \end{bmatrix}\)
\(=\begin{bmatrix}-\frac{3}{4}+2-\frac{1}{4} & \frac{1}{4}-\frac{1}{4}&\frac{3}{2}-2+\frac{1}{2} \\-\frac{3}{4}+\frac{3}{4} & \frac{1}{4}+\frac{3}{4}&\frac{3}{2}-\frac{3}{2}\\-\frac{3}{4}+1-\frac{1}{4}&\frac{1}{4}-\frac{1}{4}&\frac{3}{2}-1-\frac{1}{2} \end{bmatrix}\)
\(=\begin{bmatrix}1 & 0&1 \\0& 1&0\\0&0&1 \end{bmatrix}\)
=I
Thus, the matrix \(\begin{bmatrix}-\frac{3}{4} & \frac{1}{4}&\frac{6}{4} \\1&0&-1\\\frac{1}{4}&\frac{1}{4}&-\frac{1}{2} \end{bmatrix}\) is inverse of the given A.
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