Q.If double of Ravneet's present age is subtracted from the sum of double of Manisha's present age and Navdeep's present age then we get 4. Five years ago, if we subtract the sum of Manisha's age and Navdeep's age from the double of Ravneet's age then we get 8. After five years, if we subtract double of Manisha's age from the sum of Ravneet's and Navdeep's age then we get 1. Using matrices, find the present ages of Manisha, Ravneet and Navdeep. OR

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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