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## Answered question

2021-03-11

Use cramer's rule to determine the values of ${I}_{1},{I}_{2},{I}_{3}$ and ${I}_{4}$
$\left[\begin{array}{cccc}13.7& -4.7& -2.2& 0\\ -4.7& 15.4& 0& -8.2\\ -2.2& 0& 25.4& -22\\ 0& -8.2& -22& 31.3\end{array}\right]\left[\begin{array}{c}{I}_{1}\\ {I}_{2}\\ {I}_{3}\\ {I}_{4}\end{array}\right]=\left[\begin{array}{c}6\\ -6\\ 5\\ -9\end{array}\right]$

### Answer & Explanation

Benedict

Skilled2021-03-12Added 108 answers

Step 1
Given matrices:
$A=\left[\begin{array}{cccc}13.7& -4.7& -2.2& 0\\ -4.7& 15.4& 0& -8.2\\ -2.2& 0& 25.4& -22\\ 0& -8.2& -22& 31.3\end{array}\right],B=\left[\begin{array}{c}{I}_{1}\\ {I}_{2}\\ {I}_{3}\\ {I}_{4}\end{array}\right],C=\left[\begin{array}{c}6\\ -6\\ 5\\ -9\end{array}\right]$
Multiplication of matrices A and B is possible only when number of columns of matrix A is equal to number of rows of matrix B.
Here order of matrix A is $4×4$ and order of matrix B is $4×1$ and hence the resultant matrix C has order $4×1$
Step 2
Multiplication of matrices:
$\left[\begin{array}{cccc}13.7& -4.7& -2.2& 0\\ -4.7& 15.4& 0& -8.2\\ -2.2& 0& 25.4& -22\\ 0& -8.2& -22& 31.3\end{array}\right]\left[\begin{array}{c}{I}_{1}\\ {I}_{2}\\ {I}_{3}\\ {I}_{4}\end{array}\right]=\left[\begin{array}{c}6\\ -6\\ 5\\ -9\end{array}\right]$
$\left[\begin{array}{c}\left(13.7\right){I}_{1}-4.7{I}_{2}-2.2{I}_{3}+\left(0\right){I}_{4}\\ -4.7{I}_{1}+15.4{I}_{2}+\left(0\right){I}_{3}-8.2{I}_{4}\\ -2.2{I}_{1}+\left(0\right){I}_{2}+25.4{I}_{3}-22{I}_{4}\\ \left(0\right){I}_{1}-8.2{I}_{2}-22{I}_{3}+31.3{I}_{4}\end{array}\right]=\left[\begin{array}{c}6\\ -6\\ 5\\ -9\end{array}\right]$
From the above equation of matrix, four simultaneous equations obtained are:
$13.7{I}_{1}-4.7{I}_{2}-2.2{I}_{3}=6$
$-4.7{I}_{1}+15.4{I}_{2}-8.2{I}_{4}=-6$
$-2.2{I}_{1}+25.4{I}_{3}-22{I}_{4}=5$
$-8.2{I}_{2}-22{I}_{3}+31.3{I}_{4}=-9$
Solve the above simultaneous equations to get values of unknown variables
${I}_{1}=0.021$
${I}_{2}=-0.908$
${I}_{3}=-0.655$
${I}_{4}=-0.986$

Jeffrey Jordon

Expert2022-01-23Added 2605 answers

Answer is given below (on video)

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