Q.If $\begin{bmatrix}x&y\\x&\frac{x}{2}+t\end{bmatrix}+\begin{bmatrix}y&x+t\\x+z&\frac{x}{2}\end{bmatrix}=\begin{bmatrix}1&4\\2&3\end{bmatrix}$, find $x,y,z$ and $t$.

Showing the 12 most recent of 37 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $\begin{bmatrix} x+y \\ y+z \\ z+x \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$ then $x+y+z =$

   (a) 9

   (b) 0

   (c) 4

   (d) 5

   ›Reveal solutionSolution

   Two matrices are equal only if all corresponding entries are equal; adding all three entry-equations gives $x+y+z$ directly.

   From $\begin{bmatrix} x+y \\ y+z \\ z+x \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$, equating corresponding entries:

   …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $A = \begin{bmatrix} x & 0 \\ 2 & 3 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ given $A^2 = 9I$, then $x$ is:

   (a) $x = 4$

   (b) $x = \pm3$

   (c) $x = -3$

   (d) $x = -4$

   ›Reveal solutionSolution

   Computing $A^2$ and matching it to $9I$ forces both $x^2=9$ and $2x+6=0$; only $x=-3$ satisfies both.

   $A=\begin{bmatrix}x & 0\\ 2 & 3\end{bmatrix}$, so

   $A^2=\begin{bmatrix}x & 0\\ 2 & 3\end{bmatrix}\begin{bmatrix}x & 0\\ 2 & 3\end{bmatrix}=\begin{bmatrix}x^2 & 0\\ 2x+6 & 9\end{bmatrix}$

   …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.If [[x-2y, 0], [5, x]] = [[-3, 0], [5, 3]], then y is equal to:

   (a) 1

   (b) 3

   (c) 2

   (d) 4

   ›Reveal solutionSolution

   Two matrices are equal only if all corresponding entries are equal; comparing the $(2,2)$ entries gives $x=3$, then the $(1,1)$ entries give $y$.

   Given:

   $\begin{bmatrix}x-2y & 0\\ 5 & x\end{bmatrix} = \begin{bmatrix}-3 & 0\\ 5 & 3\end{bmatrix}$

   Comparing the $(2,2)$ entries: $x = 3$.

   …

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4. CBSE 2025Set ANNUAL1 markMCQ
   Q.For what value of $x$, $\begin{bmatrix}1&3\\2&1\end{bmatrix}\begin{bmatrix}1\\x\end{bmatrix}=\begin{bmatrix}7\\4\end{bmatrix}$?

   (i) $-2$

   (ii) $-1$

   (iii) $2$

   (iv) $1$

   ›Reveal solutionSolution

   Multiply out the matrices and compare entries.

   $\begin{bmatrix}1&3\\2&1\end{bmatrix}\begin{bmatrix}1\\x\end{bmatrix}=\begin{bmatrix}1(1)+3(x)\\2(1)+1(x)\end{bmatrix}=\begin{bmatrix}1+3x\\2+x\end{bmatrix}$

   Setting this equal to $\begin{bmatrix}7\\4\end{bmatrix}$:

   …

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5. CBSE 2025Set ANNUAL1 markMCQ
   Q.Which of the given values of $x$ and $y$ make the following pair of matrices equal? $\begin{bmatrix} 3x+7 & 5 \\ y+1 & 2-3x \end{bmatrix}, \begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix}$

   (a) $x = -\dfrac{1}{3}, y = 7$

   (b) Not possible to find

   (c) $x = -\dfrac{2}{3}, y = 7$

   (d) $x = -\dfrac{1}{3}, y = -\dfrac{2}{3}$

   ›Reveal solutionSolution

   Equating corresponding entries gives two different equations for x that contradict each other, so no consistent solution exists.

   For $\begin{bmatrix} 3x+7 & 5 \\ y+1 & 2-3x \end{bmatrix} = \begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix}$, equating each entry:

   $3x+7 = 0 \Rightarrow x = -\dfrac{7}{3}$

   $5 = y-2 \Rightarrow y = 7$

   $y+1 = 8 \Rightarrow y = 7$ (consistent with above)

   $2-3x = 4 \Rightarrow x = -\dfrac{2}{3}$

   …

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6. CBSE 2025Set ANNUAL1 markMCQ
   Q.If [[x−2y, 0], [5, x]] = [[−5, 0], [5, 3]], then y is equal to:

   (a) 1

   (b) 3

   (c) 2

   (d) 4

   ›Reveal solutionSolution

   Equal matrices have equal corresponding entries — match the (2,2) entries first to get $x$, then use the (1,1) entry to get $y$.

   Given $\begin{pmatrix} x-2y & 0 \\ 5 & x\end{pmatrix} = \begin{pmatrix} -5 & 0 \\ 5 & 3\end{pmatrix}$.

   Comparing the (2,2) entries: $x = 3$.

   …

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7. CBSE 2025Set ANNUAL1 mark
   Q.If [[a+4, 3b], [8, -14]] = [[2a+2, b+4], [8, a-8b]], then find the value of a + b.

   ›Reveal solutionSolution

   Equate corresponding entries of the two equal matrices to get $a=2$, $b=2$, so $a+b=4$.

   Two matrices are equal only if every corresponding entry is equal. Comparing entries of

   $\begin{bmatrix} a+4 & 3b \\ 8 & -14 \end{bmatrix} = \begin{bmatrix} 2a+2 & b+4 \\ 8 & a-8b \end{bmatrix}$:

   From the (1,1) entries: $a+4 = 2a+2 \Rightarrow 2 = a \Rightarrow a = 2$.

   …

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8. CBSE 2025Set ANNUAL1 markMCQ
   Q.If A = [[2x, 0], [x, x]] and A⁻¹ = [[1, 0], [−1, 2]], then x equals –

   (i) 1

   (ii) 2

   (iii) 1/2

   (iv) −2

   ›Reveal solutionSolution

   Compute $A^{-1}$ from $A=\begin{pmatrix}2x&0\\x&x\end{pmatrix}$ using the $2\times2$ inverse formula and match it to the given $A^{-1}$.

   For $A=\begin{pmatrix}2x&0\\x&x\end{pmatrix}$, $\det A = (2x)(x)-(0)(x)=2x^2$.

   Using $A^{-1}=\dfrac{1}{\det A}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ for $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$:

   $A^{-1}=\dfrac{1}{2x^2}\begin{pmatrix}x&0\\-x&2x\end{pmatrix}=\begin{pmatrix}\tfrac{1}{2x}&0\\-\tfrac{1}{2x}&\tfrac{1}{x}\end{pmatrix}$.

   …

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9. CBSE 20241 markMCQ
   Q.If $\begin{bmatrix} 8 & 14 \\ 9 & 7 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix} X$, then matrix X is : (A) $\begin{bmatrix} 3 & 7 \\ 2 & 0 \end{bmatrix}$ (B) $\begin{bmatrix} 2 & 0 \\ 7 & 3 \end{bmatrix}$ (C) $\begin{bmatrix} 2 & 0 \\ 3 & 7 \end{bmatrix}$ (D) $\begin{bmatrix} 2 & 0 \\ -3 & 7 \end{bmatrix}$

   ›Reveal solutionSolution

   We solve the matrix equation $A = BX$ by left-multiplying both sides by $B^{-1}$, giving $X = B^{-1}A$. Computing the inverse of $B = \begin{bmatrix}1&2\\3&1\end{bmatrix}$ and multiplying yields $X = \begin{bmatrix}2&0\\3&7\end{bmatrix}$, which matches option (C).

   The core idea here is that a matrix equation like $A = BX$ is solved exactly like the scalar equation $a = bx$ — you isolate $X$ by multiplying both sides by the inverse of $B$. But because matrix multiplication is not commutative, you must multiply on the left by $B^{-1}$, not on the right. That single detail is the entire key.

   Let’s walk through it.

   1. Set up the equation clearly. We are given

   $\begin{bmatrix} 8 & 14 \\ 9 & 7 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix} X.$

   Call the left matrix $A$ and the coefficient matrix $B$, so $A = BX$. Our job is to find $X$.

   2. Why left-multiplication by $B^{-1}$ works. If $B$ is invertible, then $B^{-1}B = I$, the identity matrix. Multiplying both sides of $A = BX$ on the left by $B^{-1}$ gives

   $B^{-1}A = B^{-1}(BX) = (B^{-1}B)X = IX = X.$

   So $X = B^{-1}A$. Notice: if we had multiplied on the right instead, we’d get $AB^{-1}$, which is a completely different (and wrong) matrix.

   Watch out

   A common mistake is to write $X = A B^{-1}$ by analogy with scalars. But matrix multiplication is not commutative — $B^{-1}A \neq A B^{-1}$ in general. Always multiply on the side where the inverse cancels the original matrix.

   3. Find $B^{-1}$. For a $2 \times 2$ matrix $B = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the inverse is

   $B^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix},$

   provided the determinant $ad - bc \neq 0$.

   Here $a=1$, $b=2$, $c=3$, $d=1$. The determinant is

   $\det(B) = (1)(1) - (2)(3) = 1 - 6 = -5.$

   So

   $B^{-1} = \frac{1}{-5} \begin{bmatrix} 1 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -\frac{1}{5} & \frac{2}{5} \\ \frac{3}{5} & -\frac{1}{5} \end{bmatrix}.$

   4. Multiply $B^{-1}A$. Now $A = \begin{bmatrix} 8 & 14 \\ 9 & 7 \end{bmatrix}$. Compute $X = B^{-1}A$:

   $X = \begin{bmatrix} -\frac{1}{5} & \frac{2}{5} \\ \frac{3}{5} & -\frac{1}{5} \end{bmatrix} \begin{bmatrix} 8 & 14 \\ 9 & 7 \end{bmatrix}.$

   Multiply entry by entry: …

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10. CBSE 2024Set D1 markMCQ
    Q.If $2A + B + X = 0$, where $A = \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & -2 \\ 1 & 5 \end{bmatrix}$ then $X =$

    (a) $\begin{bmatrix} 1 & 2 \\ -7 & -13 \end{bmatrix}$

    (b) $\begin{bmatrix} 1 & 2 \\ 7 & 13 \end{bmatrix}$

    (c) $\begin{bmatrix} -1 & -2 \\ -7 & -13 \end{bmatrix}$

    (d) $\begin{bmatrix} -1 & -2 \\ 7 & 13 \end{bmatrix}$

    ›Reveal solutionSolution

    From $2A + B + X = 0$, solve $X = -2A - B$.

    $2A = \begin{bmatrix} -2 & 4 \\ 6 & 8 \end{bmatrix}$, so …

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11. CBSE 2024Set D1 markMCQ
    Q.$[\,x\ \ y\,] = [\,2x - 1\ \ 9\,] \Rightarrow$

    (a) $x = 3,\ y = 9$

    (b) $x = 1,\ y = 9$

    (c) $x = 0,\ y = 9$

    (d) $x = 3,\ y = 4$

    ›Reveal solutionSolution

    Equal matrices have equal corresponding entries.

    …

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12. CBSE 2024Set ANNUAL1 markMCQ
    Q.If $\begin{bmatrix} x+y+z \\ x+z \\ y+z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}$ then $x+y+z =$

    (a) $5$

    (b) $7$

    (c) $9$

    (d) none of these

    ›Reveal solutionSolution

    Matching the first row of the given matrix equation reads off x+y+z directly, no further algebra needed.

    The matrix equation $\begin{bmatrix}x+y+z\\x+z\\y+z\end{bmatrix}=\begin{bmatrix}9\\5\\7\end{bmatrix}$ means corresponding entries are equal:

    Row 1: $x+y+z=9$ …

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