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Solve AX = B by the matrix method

This type of question, asked 8 times: 2 May 2023 S2, 3 May 2023 S2, 11 May 2023 S1, 9 May 2024 S2, 20 Apr 2025 S2, 11 Apr 2026 S2, 13 May 2026 S1, 15 May 2026 S1. All 8 questions below, oldest first.

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  1. Question 1

    2 May 2023 S2

    102
    Let X=(abc)X = \begin{pmatrix}a\\b\\c\end{pmatrix}, A=(1−12201321)A = \begin{pmatrix}1&-1&2\\2&0&1\\3&2&1\end{pmatrix} and B=(314)B = \begin{pmatrix}3\\1\\4\end{pmatrix}. If AX=BAX = B, then the value of 2a−3b+4c2a - 3b + 4c will be

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  2. Question 2

    3 May 2023 S2

    120
    For the system x−y+z=4,  2x+y−3z=0,  x+y+z=2x-y+z=4,\;2x+y-3z=0,\;x+y+z=2, the values of x,y,zx,y,z respectively are given by

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  3. Question 3

    11 May 2023 S1

    131
    Let A=(1−1121−3111)A = \begin{pmatrix}1&-1&1\\2&1&-3\\1&1&1\end{pmatrix} and B=(402)B = \begin{pmatrix}4\\0\\2\end{pmatrix} such that AX=BAX = B, then X=X =

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  4. Question 4

    9 May 2024 S2

    143
    Let A=[1110131−21], B=[6110]A=\begin{bmatrix}1&1&1\\0&1&3\\1&-2&1\end{bmatrix},\ B=\begin{bmatrix}6\\11\\0\end{bmatrix} and X=[abc]X=\begin{bmatrix}a\\b\\c\end{bmatrix}, if AX=BAX=B, then the value of 2a+b+2c2a+b+2c is

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  5. Question 5

    20 Apr 2025 S2

    127
    If [133144134][xyz]=[121513]\begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 4 \\ 1 & 3 & 4 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 12 \\ 15 \\ 13 \end{bmatrix}, then the value of x2+y2+z2=x^{2}+y^{2}+z^{2}=

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  6. Question 6

    11 Apr 2026 S2

    66
    If [21−1−1211−12][abc]=[40−2]\begin{bmatrix} 2 & 1 & -1 \\ -1 & 2 & 1 \\ 1 & -1 & 2 \end{bmatrix}\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 4 \\ 0 \\ -2 \end{bmatrix} then the distance of point P(a,b,c)P(a, b, c) from the plane 2x+y+2z=102x + y + 2z = 10 is

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  7. Question 7

    13 May 2026 S1

    69
    Let AT=[1121−1112−1]A^T = \begin{bmatrix} 1 & 1 & 2 \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{bmatrix}, BT=[651]B^T = \begin{bmatrix} 6 & 5 & 1 \end{bmatrix} and X=[xyz]X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}. If AX=BAX = B, then the values of x,y,zx, y, z are..........

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  8. Question 8

    15 May 2026 S1

    70
    If [1112−17312][xyz]=[172]\begin{bmatrix} 1 & 1 & 1 \\ 2 & -1 & 7 \\ 3 & 1 & 2 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 7 \\ 2 \end{bmatrix} then [xyz]=\begin{bmatrix} x \\ y \\ z \end{bmatrix} =

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