Solve the following system of equations, using matrices: $\dfrac{2}{x} + \dfrac{3}{y} + \dfrac{10}{z} = 4,\ \dfrac{4}{x} - \dfrac{6}{y} + \dfrac{5}{z} = 1,\ \dfrac{6}{x} + \dfrac{9}{y} - \dfrac{20}{z} = 2$ where $x, y, z \ne 0$. OR If $A = \begin{bmatrix} 1 & \cot x \\ -\cot x & 1 \end{bmatrix}$, show that $A'A^{-1} = \begin{bmatrix} -\cos 2x & -\sin 2x \\ \sin 2x & -\cos 2x \end{bmatrix}$.

Solve the following system of equations, using matrices — CBSE Class 12 Mathematics 2024 PYQ

Question

\dfrac{2}{x} + \dfrac{3}{y} + \dfrac{10}{z} = 4,\ \dfrac{4}{x} - \dfrac{6}{y} + \dfrac{5}{z} = 1,\ \dfrac{6}{x} + \dfrac{9}{y} - \dfrac{20}{z} = 2

(OR)

A = \begin{bmatrix} 1 & \cot x \\ -\cot x & 1 \end{bmatrix}

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Part (a)Concept understanding — Inverse Matrix Method

The Inverse Matrix Method

Many problems reduce to a system of linear equations, for example

\begin{aligned}2x+3y&=8\\x-\ \ y&=-1.\end{aligned}

The inverse matrix method solves such a system by writing it as a single matrix equation and then undoing the coefficient matrix with its inverse — the matrix analogue of dividing.

Writing the system as $AX=B$

Collect the coefficients, the unknowns and the constants:

A=\begin{pmatrix}2&3\\1&-1\end{pmatrix},\quad X=\begin{pmatrix}x\\y\end{pmatrix},\quad B=\begin{pmatrix}8\\-1\end{pmatrix},

so the whole system becomes $AX=B$ .

The idea

For numbers, $ax=b$ gives $x=a^{-1}b$ provided $a\neq0$ . The same works for matrices: if $A$ is invertible, multiply $AX=B$ on the left by $A^{-1}$ :

A^{-1}(AX)=A^{-1}B\;\Rightarrow\;IX=A^{-1}B\;\Rightarrow\;X=A^{-1}B.

X=A^{-1}B

Multiplying on the left matters — matrix products do not commute, so $BA^{-1}$ would be wrong.

When it works

The inverse $A^{-1}$ exists only when $\det A\neq0$ , so:

$\det A\neq0$ : the system is consistent with the unique solution $X=A^{-1}B$ .
$\det A=0$ : no inverse; the system is either inconsistent (no solution) or has infinitely many — handle it by another method.

Worked steps

For the system above, $\det A=(2)(-1)-(3)(1)=-5\neq0$ , and

A^{-1}=\frac{1}{-5}\begin{pmatrix}-1&-3\\-1&2\end{pmatrix}.

Then …

Part (b)Concept understanding — Inverse Matrix Method

The Inverse Matrix Method

Many problems reduce to a system of linear equations, for example

\begin{aligned}2x+3y&=8\\x-\ \ y&=-1.\end{aligned}

The inverse matrix method solves such a system by writing it as a single matrix equation and then undoing the coefficient matrix with its inverse — the matrix analogue of dividing.

Writing the system as $AX=B$

Collect the coefficients, the unknowns and the constants:

A=\begin{pmatrix}2&3\\1&-1\end{pmatrix},\quad X=\begin{pmatrix}x\\y\end{pmatrix},\quad B=\begin{pmatrix}8\\-1\end{pmatrix},

so the whole system becomes $AX=B$ .

The idea

For numbers, $ax=b$ gives $x=a^{-1}b$ provided $a\neq0$ . The same works for matrices: if $A$ is invertible, multiply $AX=B$ on the left by $A^{-1}$ :

A^{-1}(AX)=A^{-1}B\;\Rightarrow\;IX=A^{-1}B\;\Rightarrow\;X=A^{-1}B.

X=A^{-1}B

Multiplying on the left matters — matrix products do not commute, so $BA^{-1}$ would be wrong.

When it works

The inverse $A^{-1}$ exists only when $\det A\neq0$ , so:

$\det A\neq0$ : the system is consistent with the unique solution $X=A^{-1}B$ .
$\det A=0$ : no inverse; the system is either inconsistent (no solution) or has infinitely many — handle it by another method.

Worked steps

For the system above, $\det A=(2)(-1)-(3)(1)=-5\neq0$ , and

A^{-1}=\frac{1}{-5}\begin{pmatrix}-1&-3\\-1&2\end{pmatrix}.

Then …

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Part (a): the system gives $x=2,\ y=3,\ z=5$ . Part (b): computing $A'$ and $A^{-1}$ and multiplying gives $A'A^{-1}=\begin{bmatrix}-\cos2x&-\sin2x\\\sin2x&-\cos2x\end{bmatrix}$ .

Part (a)

The equations are linear not in $x,y,z$ but in their reciprocals, so we substitute $u=\dfrac1x,\ v=\dfrac1y,\ w=\dfrac1z$ (valid since $x,y,z\neq0$ ):

2u+3v+10w=4,\quad 4u-6v+5w=1,\quad 6u+9v-20w=2.

1. Matrix form $AU=B$ .

A=\begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20\end{bmatrix},\quad U=\begin{bmatrix}u\\v\\w\end{bmatrix},\quad B=\begin{bmatrix}4\\1\\2\end{bmatrix}.

2. Determinant.

\det A = 2(120-45)-3(-80-30)+10(36+36)=150+330+720=1200\neq0,

so $A^{-1}$ exists and the solution is unique.

3. Adjoint and inverse. The cofactor matrix transposes to

\operatorname{adj}A=\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix},\qquad A^{-1}=\frac{1}{1200}\operatorname{adj}A.

4. Solve $U=A^{-1}B$ .

U=\frac{1}{1200}\begin{bmatrix}300+150+150\\440-100+60\\288+0-48\end{bmatrix}=\frac{1}{1200}\begin{bmatrix}600\\400\\240\end{bmatrix}=\begin{bmatrix}1/2\\1/3\\1/5\end{bmatrix}.

5. Back-substitute. $x=1/u=2,\ y=1/v=3,\ z=1/w=5$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$A$
(a) $\det(A)$
(b) $\frac{1}{\det(A)}$
(c) $1$
(d) $0$
›Reveal solutionSolution

$\det(A^{-1})=\dfrac{1}{\det A}$ .

From $AA^{-1}=I$ , $\det(A)\det(A^{-1})=\det(I)=1$ , so $\det(A^{-1})=\dfrac{1}{\det(A)}$ . (This holds for a …

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CBSE 2026Set ANNUAL1 mark
$AX = B$
›Reveal solutionSolution

The unique solution of $AX=B$ (when $|A| e0$ ) is $X=A^{-1}B$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$
(a) $19$
(b) $\dfrac{1}{19}$
(c) $-19$
(d) $-\dfrac{1}{19}$
›Reveal solutionSolution

Compute |A| and adj(A), form A⁻¹, and compare it entry-by-entry with kA.

$A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$ , $|A| = 2(-2) - 3(5) = -4-15 = -19$

$\text{adj}(A) = \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}$

…

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CBSE 2024Set D1 markMCQ
$A = \begin{bmatrix} 2 & -3 \\ 4 & 6 \end{bmatrix}$
(a) $\begin{bmatrix} \frac{1}{4} & \frac{1}{8} \\ \frac{1}{6} & \frac{1}{12} \end{bmatrix}$
(b) $\begin{bmatrix} \frac{1}{4} & \frac{1}{8} \\ -\frac{1}{6} & \frac{1}{12} \end{bmatrix}$
(c) $\begin{bmatrix} 4 & 8 \\ 6 & 12 \end{bmatrix}$
(d) $\begin{bmatrix} 4 & 8 \\ -6 & 12 \end{bmatrix}$
›Reveal solutionSolution

For a $2\times2$ matrix $A^{-1} = \frac{1}{\det A}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ .

Here $A = \begin{bmatrix} 2 & -3 \\ 4 & 6 \end{bmatrix}$ , so $\det A = (2)(6) - (-3)(4) = 12 + 12 = 24.$

The adjoint (swap diagonal, negate off-diagonal) is $\begin{bmatrix} 6 & 3 \\ -4 & 2 \end{bmatrix}$ . Hence …

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Part (a)

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