# 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

Let $A=\left[\begin{array}{ccc}1& 2& -1\\ 1& 0& 3\\ 1& 1& -1\end{array}\right]$
i) Find the determinant of A
ii)Verify that the matrix
$\left[\begin{array}{ccc}-\frac{3}{4}& \frac{1}{4}& \frac{6}{4}\\ 1& 0& -1\\ \frac{1}{4}& \frac{1}{4}& -\frac{1}{2}\end{array}\right]$ is the inverse matrix of A
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Step 1
For part (i) of a:
According to the given information, it is required to calculate the determinant of matrix
$A=\left[\begin{array}{ccc}1& 2& -1\\ 1& 0& 3\\ 1& 1& -1\end{array}\right]$
For this use the formula
$|\begin{array}{ccc}{x}_{1}& {y}_{1}& {z}_{1}\\ {x}_{2}& {y}_{2}& {z}_{2}\\ {x}_{3}& {y}_{3}& {z}_{3}\end{array}|={x}_{1}|\begin{array}{cc}{y}_{2}& {z}_{2}\\ {y}_{3}& {z}_{3}\end{array}|-{y}_{1}|\begin{array}{cc}{x}_{2}& {z}_{2}\\ {x}_{3}& {z}_{3}\end{array}|+{z}_{1}|\begin{array}{cc}{x}_{2}& {y}_{2}\\ {x}_{3}& {y}_{3}\end{array}|$
Step 2
Using above formula determinant of A can be calculated as
$|A|=|\begin{array}{ccc}1& 2& -1\\ 1& 0& 3\\ 1& 1& -1\end{array}|$
$=1|\begin{array}{cc}0& 3\\ 1& -1\end{array}|-2|\begin{array}{cc}1& 3\\ 1& -1\end{array}|-1|\begin{array}{cc}1& 0\\ 1& 1\end{array}|$
$=1\left(0×-1-3×1\right)-2\left(1×-1-3×1\right)-1\left(1×1-0×1\right)$
=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
$\left[\begin{array}{ccc}-\frac{3}{4}& \frac{1}{4}& \frac{6}{4}\\ 1& 0& -1\\ \frac{1}{4}& \frac{1}{4}& -\frac{1}{2}\end{array}\right]$
is inverse of given matrix A use the definition of identity matrix that states:
$A{A}^{-1}=I$
Step 4
Now calculate the multiplication of these two matrices, as
$\left[\begin{array}{ccc}1& 2& -1\\ 1& 0& 3\\ 1& 1& -1\end{array}\right]\left[\begin{array}{ccc}-\frac{3}{4}& \frac{1}{4}& \frac{6}{4}\\ 1& 0& -1\\ \frac{1}{4}& \frac{1}{4}& -\frac{1}{2}\end{array}\right]=\left[\begin{array}{ccc}1×-\frac{3}{4}+2×1-1×\frac{1}{4}& 1×\frac{1}{4}+2×0-1×\frac{1}{4}& 1×\frac{6}{4}+2×-1-1×-\frac{1}{2}\\ 1×-\frac{3}{4}+0×1+3×\frac{1}{4}& 1×\frac{1}{4}+0×0+3×\frac{1}{4}& 1×\frac{6}{4}+0×-1+3×-\frac{1}{2}\\ 1×-\frac{3}{4}+1×1-1×\frac{1}{4}& 1×\frac{1}{4}+1×0-1×\frac{1}{4}& 1×\frac{6}{4}+1×-1-1×-\frac{1}{2}\end{array}\right]$
Jeffrey Jordon