Chapter 10.5, Problem 2E

(a)

To determine

To express: The given system of equations as a matrix equation.

Answer to Problem 2E

The given system of equations as a matrix equation is $\left[\begin{array}{cc}5& 3\\ 3& 2\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}4\\ 3\end{array}\right]$.

Explanation of Solution

The given system of equations is,

$\{\begin{array}{l}5x+3y=4\\ 3x+2y=3\end{array}$

Coefficient matrix A is,

$A=\left[\begin{array}{cc}5& 3\\ 3& 2\end{array}\right]$.

Variable matrix X is,

$X=\left[\begin{array}{c}x\\ y\end{array}\right]$.

Known matrix B is,

$B=\left[\begin{array}{c}4\\ 3\end{array}\right]$.

The given matrix equation is,

$A\cdot X=B$.

Thus, the given system of equations as a matrix equation is $\left[\begin{array}{cc}5& 3\\ 3& 2\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}4\\ 3\end{array}\right]$.

(b)

To determine

To find: The inverse of matrix A.

Answer to Problem 2E

The inverse of matrix A is $\left[\begin{array}{cc}2& -3\\ -3& 5\end{array}\right]$.

Explanation of Solution

Given:

The matrix A from part (a) is,

$A=\left[\begin{array}{cc}5& 3\\ 3& 2\end{array}\right]$.

Calculation:

Take a $2\times 4$ matrix with left two columns is given matrix and right two columns is identity matrix to find the inverse of matrix A.

$\left[\begin{array}{ccccc}5& 3& & 1& 0\\ 3& 2& & 0& 1\end{array}\right]$

Apply elementary row operations on the above matrix and convert the left half of matrix into the identity matrix.

Apply $\left({\text{R}}_1\to {\text{R}}_1-{\text{R}}_2\right)$ in the above matrix.

$\left[\begin{array}{ccccc}2& 1& & 1& -1\\ 3& 2& & 0& 1\end{array}\right]$

Apply $\left({\text{R}}_2\to {\text{R}}_2-{\text{R}}_1\right)$ in the above matrix.

$\left[\begin{array}{ccccc}2& 1& & 1& -1\\ 1& 1& & -1& 2\end{array}\right]$

Apply $\left({\text{R}}_1\to {\text{R}}_1-{\text{R}}_2\right)$ in the above matrix.

$\left[\begin{array}{ccccc}1& 0& & 2& -3\\ 1& 1& & -1& 2\end{array}\right]$

Apply $\left({\text{R}}_2\to {\text{R}}_2-{\text{R}}_1\right)$ in the above matrix.

$\left[\begin{array}{ccccc}1& 0& & 2& -3\\ 0& 1& & -3& 5\end{array}\right]$

The right half of the above matrix is the inverse of matrix.

Thus, the inverse of matrix A is $\left[\begin{array}{cc}2& -3\\ -3& 5\end{array}\right]$.

(c)

To determine

To find: The solution of matrix equation $X=A^{-1}B$.

Answer to Problem 2E

The solution of matrix equation is $\left[\begin{array}{c}-1\\ 3\end{array}\right]$.

Explanation of Solution

The inverse of matrix A from part (b) is,

$A^{-1}=\left[\begin{array}{cc}2& -3\\ -3& 5\end{array}\right]$

The matrix B from part A is,

$B=\left[\begin{array}{c}4\\ 3\end{array}\right]$

The matrix equation is,

$\begin{array}{c}X=A^{-1}B\\ \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{cc}2& -3\\ -3& 5\end{array}\right]\left[\begin{array}{c}4\\ 3\end{array}\right]\\ =\left[\begin{array}{c}8-9\\ -12+15\end{array}\right]\\ =\left[\begin{array}{c}-1\\ 3\end{array}\right]\end{array}$

Thus, the solution of matrix equation is $\left[\begin{array}{c}-1\\ 3\end{array}\right]$.

(d)

To determine

To fill: The value of x and y.

Answer to Problem 2E

The value of x is $-1$ and y is 3.

Explanation of Solution

The solution of system from part (c) is,

$\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-1\\ 3\end{array}\right]$

Compare row 1 and row 2 on both sides.

$x=-1$ and $y=3$

Thus, the value of x is $-1$ and y is 3.