Question:
If \( \begin{pmatrix} 1 & 2 & -3 \\ 0 & 4 & 5 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \), then \( (x, y, z) \) is equal to:
If \( \begin{pmatrix} 1 & 2 & -3 \\ 0 & 4 & 5 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \), then \( (x, y, z) \) is equal to:
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Always check the diagonal of the matrix. If all entries below the diagonal are zero, start solving from the last variable. It avoids the need for complex matrix inversion.
Updated On: May 1, 2026
- \( (1, 6, 6) \)
- \( (1, -6, 1) \)
- \( (1, 1, 6) \)
- \( (6, -1, 1) \)
- \( (-1, 6, 1) \)
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The Correct Option is D
Solution and Explanation
Concept: The given matrix is an upper triangular matrix. For such systems, the method of back-substitution is the most efficient way to find the variables, starting from the bottom row and working upwards.
Step 1: Extracting equations from the matrix multiplication.
From the third row: \[ 1z = 1 \implies z = 1 \] From the second row: \[ 4y + 5z = 1 \] From the first row: \[ x + 2y - 3z = 1 \]
Step 2: Back-substituting to find \( y \) and \( x \).
Using \( z = 1 \) in the second equation: \[ 4y + 5(1) = 1 \implies 4y = -4 \implies y = -1 \] Now, using \( y = -1 \) and \( z = 1 \) in the first equation: \[ x + 2(-1) - 3(1) = 1 \] \[ x - 2 - 3 = 1 \implies x - 5 = 1 \implies x = 6 \] The solution set is \( (x, y, z) = (6, -1, 1) \).
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