Question

Match each system of equations to the inverse of its coefficient matrix, A-1, and the matrix of its solution, X.

Answer 1

The system of equations to the inverse of its coefficient matrix, A⁻¹, and the matrix of its solution, X is shown in the figure.

Given that the system of equations are shown in given figure.

The first system of equations are

$\begin{aligned}4x+2y-z&=150\\x+y-z&=-100\\-3x-y+z&=600\\\end$

By writing in matrix AX=b, we get

Coefficient matrix $A=\left[\begin{array}{lll}4&2&-1\\1&1&-1\\-3&-1&1\end{array}\right]$ and $B=\left[\begin{array}{l}150&-100&600\end{array}\right]$

Firstly, we will find the A⁻¹ by finding the determinant and adjoint of A and divide the adjoint with determinant, we get

$\begin{aligned}|A|&=\left|\begin{array}{lll}4&2&-1\\1&1&-1\\-3&-1&1\end{array}\right|\\ &=4(1-1)-2(1-3)-1(-1+3)\\&=4(0)-2(-2)-1(2)\\ &=2\neq 0\end$

$\begin{aligned}Adj A&=\left[\begin{array}{lll}0&2&2\\-1&1&-2\\-2&3&2\end{array}\right]^T\\&=\left[\begin{array}{lll}0&-1&-2\\2&1&3\\2&-2&2\end{array}\right]\end$

$\begin{aligned}A^{-1}&=\frac{Adj A}{|A|}\\ &=\left[\begin{array}{lll}0&-0.5&-0.5\\1&0.5&1.5\\1&-1&1\end{array}\right]\end$

For a solution Consider [A B] and apply row operations, we get

$\begin{aligned}\left[A\right.\text{ }\left.B\right]&=\left[\begin{array}{lll1}4&2&-1&150\\1&1&-1&-100\\-3&-1&1&600\end{array}\right]\\ R_{2}&\rightarrow 4R_{2}-R_{1},R_{3}\rightarrow 4R_{3}+3R_{1}\\ &\sim \left[\begin{array}{lll1}4&2&-1&150\\0&2&-3&-550\\0&2&1&2850\end{array}\right]\\ R_{3}&\rightarrow R_{3}-R_{2}\\ &\sim \left[\begin{array}{llll}4&2&-1&150\\0&2&-3&-550\\0&0&4&3400\end{array}\right]\end$

Thus, $x=\left[\begin{array}{l}x\\y\\z\end{array}\right]=\left[\begin{array}{l}-250\\1000\\850\end{array}\right]$

The second system of equations are

$\begin{aligned}x+y-z&=220\\5x-5y-z&=-640\\-x+y+z&=200\\\end$

Similarly, we will find for second system of equations

$\begin{aligned}|A|&=\left|\begin{array}{lll}1&1&-1\\5&-5&-1\\-1&1&1\end{array}\right|\\ &=1(-5+1)-1(5-1)-1(5-5)\\&=1(-4)-1(4)-1(0)\\ &=-8\neq 0\end$

$\begin{aligned}Adj A&=\left[\begin{array}{lll}-4&-4&0\\-2&0&-2\\-6&-4&-10\end{array}\right]^T\\&=\left[\begin{array}{lll}-4&-2&-6\\-4&0&-4\\0&-2&-10\end{array}\right]\end$

$\begin{aligned}A^{-1}&=\frac{Adj A}{|A|}\\ &=\left[\begin{array}{lll}0.5&0.25&0.75\\0.5&0&0.5\\0&0.25&1.25\end{array}\right]\end$

$\begin{aligned}\left[A\right.\text{ }\left.B\right]&=\left[\begin{array}{llll}1&1&-1&220\\5&-5&-1&-640\\-1&1&1&200\end{array}\right]\\ R_{2}&\rightarrow R_{2}-5R_{1},R_{3}\rightarrow R_{3}+R_{1}\\ &\sim \left[\begin{array}{llll}1&1&-1&220\\0&-10&4&-1740\\0&2&0&420\end{array}\right]\\ R_{3}&\rightarrow 5R_{3}+R_{2}\\ &\sim \left[\begin{array}{llll}1&1&-1&220\\0&-10&4&-1740\\0&0&4&360\end{array}\right]\end$

Thus, $x=\left[\begin{array}{l}x\\y\\z\end{array}\right]=\left[\begin{array}{l}100\\210\\90\end{array}\right]$

The third system of equations are

$\begin{aligned}2x+2y-z&=290\\x+y-3z&=500\\x-y+2z&=600\\\end$

Similarly, we will find for third system of equations

$\begin{aligned}|A|&=\left|\begin{array}{lll}2&2&-1\\1&1&-3\\1&-1&2\end{array}\right|\\ &=2(2-3)-2(2+3)-1(-1-1)\\&=2(-1)-2(5)-1(-2)\\ &=-10\neq 0\end$

$\begin{aligned}Adj A&=\left[\begin{array}{lll}-1&-5&-2\\-3&5&4\\-5&5&0\end{array}\right]^T\\&=\left[\begin{array}{lll}-1&-3&-5\\-5&5&5\\-2&4&0\end{array}\right]\end$

$\begin{aligned}A^{-1}&=\frac{Adj A}{|A|}\\ &=\left[\begin{array}{lll}0.1&0.3&0.5\\0.5&-0.5&-0.5\\0.2&-0.4&0\end{array}\right]\end$

get

$\begin{aligned}\left[A\right.\text{ }\left.B\right]&=\left[\begin{array}{llll}2&2&-1&290\\1&1&-3&500\\1&-1&2&600\end{array}\right]\\ R_{2}&\rightarrow 2R_{2}-R_{1},R_{3}\rightarrow 2R_{3}-R_{1}\\ &\sim \left[\begin{array}{llll}2&2&-1&290\\&0&-5&710\\0&-4&5&910\end{array}\right]\end$

Thus, $x=\left[\begin{array}{l}x\\y\\z\end{array}\right]=\left[\begin{array}{l}479\\-405\\-142\end{array}\right]$

Hence, each system of equations to the inverse of its coefficient matrix, A⁻¹, and the matrix of its solution, X.

Learn more about the system of equations from here brainly.com/question/1761142

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