316-60-1608-17

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316-60-1608-17

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To solve the matrix equation, let&#039;s start by writing the given matrix:A=[316−601608−17]To find the solution, we&#039;ll use the method of Gaussian elimination. The goal is to transform the matrix into a reduced row-echelon form, where the leading coefficient in each row is 1 and all other entries in the column are 0.Step 1: Interchange rows if necessary to bring a nonzero entry to the top of the first column. In this case, the first entry in the first column is already nonzero, so we proceed to step 2.Step 2: Use row operations to create zeros below the first entry in the first column. We can do this by multiplying the first row by 2 and adding it to the second row:R2←R2+2R1[316022808−17]Step 3: Use row operations to create zeros below the second entry in the second column. We can achieve this by multiplying the second row by 4 and subtracting 8 times the second row from the third row:R3←R3−4R2[316022800−115]Now, we have the matrix in row-echelon form. Let&#039;s proceed to step 4.Step 4: Use row operations to create leading coefficients of 1. We can accomplish this by dividing the second row by 2 and the third row by -115:R2←12R2R3←1−115R3[3160114001]Step 5: Use row operations to create zeros above each leading coefficient of 1. We can achieve this by subtracting the first row from the second row, and then subtracting 14 times the second row from the third row:R2←R2−R1R3←R3−14R2[3160114001]Step 6: Finally, use row operations to create zeros above each leading coefficient of 1. Subtract 6 times the third row from the first row and subtract 14 times the third row from the second row:R1←R1−6R3R2←R2−14R3[310010001]We have obtained the reduced row-echelon form of the matrix. The solution to the matrix equation is:x=0y=0z=1Therefore, the solution to the matrix equation is x=0, y=0, and z=1.

316-60-1608-17

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