(a) If $P = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}$ and $Q = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}$, find $(QP)$ and hence solve the following system of equations using matrices: $x - y = 3,\ 2x + 3y + 4z = 17,\ y + 2z = 7$ OR (b) Obtain the value of $\Delta = \begin{vmatrix} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{vmatrix}$ in terms of $x, y$ and $z$. Further, if $\Delta = 0$ and $x, y, z$ are non-zero real numbers, prove that $x^{-1} + y^{-1} + z^{-1} = -1$.

(a) If P = 1 -1 0 \2 3 4 \0 1 2 and Q = 2 2 -4 \-4 2 -4 \2 -1 5 , find (QP) and hence solve the following syst

Question

P = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}

(OR)

(b) Obtain the value of

\Delta = \begin{vmatrix} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{vmatrix}

in terms of

x, y

and

z

. Further, if

\Delta = 0

and

x, y, z

are non-zero real numbers, prove that

x^{-1} + y^{-1} + z^{-1} = -1

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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 — Determinant Evaluation Using Identities

Determinant Evaluation Using Identities

Expanding a $4\times4$ or $5\times5$ determinant term by term is painful and error-prone. The smarter route is to transform the determinant into an easy form using properties (the "identities") that change its value in a known, controlled way — then read the answer off a triangular matrix.

The geometric intuition

A determinant measures the signed "volume" of the box spanned by the rows in $n$ -dimensional space. Sliding one row parallel to another doesn't change that volume; swapping two rows flips its sign; scaling a row scales the volume. The algebraic identities are just these facts translated into rules.

The three row (or column) operations

\det \to -\det

The identical rules hold for columns. There is also row-wise linearity: if a row is a sum $R_i = R_i' + R_i''$ , the determinant splits into the sum of two determinants with all other rows fixed.

Watch out

Row-wise linearity is not $\det(A+B)=\det A+\det B$ — that is false. The splitting works one row at a time.

The strategy

Use operation 3 to create zeros in a row or column (value unchanged).
Factor out common factors with operation 2.
Swap rows if needed to reach upper-triangular form (track the sign change).
The determinant is then the product of the diagonal entries.

Worked example

\det\begin{pmatrix}1&2&3\\4&5&6\\7&8&10\end{pmatrix}.

Apply $R_2\to R_2-4R_1$ and $R_3\to R_3-7R_1$ (no change), then $R_3\to R_3-2R_2$ : …

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(a) $QP=6I$ , so $P^{-1}=\frac16 Q$ and the system $PX=B$ gives $x=2,\ y=-1,\ z=4$ . (b) $\Delta=xyz\left(1+\frac1x+\frac1y+\frac1z\right)$ ; setting $\Delta=0$ with non-zero $x,y,z$ gives $x^{-1}+y^{-1}+z^{-1}=-1$ .

Part (a) — $QP$ and solving the system

Step 1: Compute $QP$ . Entry-by-entry,

QP=\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}=\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}=6I.

Step 2: Read off $P^{-1}$ . From $QP=6I$ we get $\left(\tfrac16 Q\right)P=I$ , so $P^{-1}=\tfrac16 Q$ .

Step 3: Set up the system in matrix form. The equations $x-y=3,\ 2x+3y+4z=17,\ y+2z=7$ are $PX=B$ with

P=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}3\\17\\7\end{bmatrix}.

Step 4: Solve. $X=P^{-1}B=\tfrac16 QB$ :

$X=\frac16\begin{bmatrix}2(3)+2(17)-4(7)\\-4(3)+2(17)-4(7)\\2(3)-1(17)+5(7)\end{bmatrix}=\frac16\begin{bmatrix}6+34-28\\-12+34-28\\6-17+35\end{bmatrix}=\frac16\begin{bmatrix}12\\-6\\24\end{bmatrix}=\begin{bmatrix}2\\-1\\4\end{bmatrix}.$ …

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

Showing the 12 most recent of 63 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$\Delta_1 = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{vmatrix}$
›Reveal solutionSolution

The first determinant is diagonal; the second requires one row interchange to reach diagonal form. Each interchange flips the sign; evaluating both determinants gives $\Delta_2 = -2\Delta_1$ . The answer is (B).

Why determinants change under row operations

A determinant measures the signed volume of the parallelepiped spanned by the row vectors. When you swap two rows, you reflect the figure across a hyperplane—the volume stays the same in magnitude but the orientation reverses, flipping the sign.

The diagonal determinant is the easiest to compute: the product of the diagonal entries. The second determinant looks scrambled, but a single row swap will bring it into a form we recognize.

Step-by-step evaluation

1. Compute $\Delta_1$ directly.

The matrix is diagonal:
$\Delta_1 = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{vmatrix} = 1 \cdot 2 \cdot 3 = 6.$
2. Recognize the structure of $\Delta_2$ .
$\Delta_2 = \begin{vmatrix} 0 & 2 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 6 \end{vmatrix}.$
The first two rows are out of order compared to a diagonal form. Swap rows 1 and 2 to bring the $1$ into the top-left position.

3. Apply the row-interchange property.

Swapping rows 1 and 2:
$\Delta_2 = -\begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 6 \end{vmatrix}.$
The negative sign comes from the single interchange.

4. Evaluate the new diagonal determinant.
$\begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 6 \end{vmatrix} = 1 \cdot 2 \cdot 6 = 12.$
So $\Delta_2 = -12$ . …
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CBSE 2026Set A1 markMCQ
$\begin{vmatrix} 23 & 12 & 11 \\ 36 & 10 & 26 \\ 63 & 26 & 37 \end{vmatrix} =$
(a) $1$
(b) $-1$
(c) $0$
(d) $2$
›Reveal solutionSolution

The determinant equals $0$ because one column is the sum of the other two.

Inspect the columns of

$\begin{vmatrix} 23 & 12 & 11 \\ 36 & 10 & 26 \\ 63 & 26 & 37 \end{vmatrix}.$

Check: $12+11 = 23$ , $10+26 = 36$ , $26+37 = 63$ .

…

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CBSE 2026Set A1 markMCQ
$\begin{vmatrix} \cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ \end{vmatrix} =$
(a) $1$
(b) $0$
(c) $-1$
(d) $\frac{1}{2}$
›Reveal solutionSolution

The determinant equals $\cos 90^\circ = 0$ .

Expand:

$\begin{vmatrix} \cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ \end{vmatrix} = \cos 15^\circ\cos 75^\circ - \sin 15^\circ\sin 75^\circ.$

…

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CBSE 2026Set A1 markMCQ
$\begin{vmatrix} a+ib & c+id \\ -c+id & a-ib \end{vmatrix} =$
(a) $a^2 + b^2 + c^2 + d^2$
(b) $a^2 - b^2 - c^2 - d^2$
(c) $a^2 - b^2 + c^2 + d^2$
(d) $a^2 + b^2 + c^2 - d^2$
›Reveal solutionSolution

Expand the $2\times2$ determinant and simplify the complex products.

$\begin{vmatrix} a+ib & c+id \\ -c+id & a-ib \end{vmatrix} = (a+ib)(a-ib) - (c+id)(-c+id).$

First term: $(a+ib)(a-ib) = a^2 - (ib)^2 = a^2 + b^2$ . …

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CBSE 2026Set ANNUAL1 markMCQ
$\begin{vmatrix}x^2-x+1 & x-1\\ x+1 & x+1\end{vmatrix}$
(a) $x^2-x+2$
(b) $x^3+x^2-2$
(c) $x^3-x^2+2$
(d) $x^3+x^2+4$
›Reveal solutionSolution

Expand the $2\times2$ determinant using $\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc$ .

$\Delta = (x^2-x+1)(x+1) - (x-1)(x+1)$

$(x^2-x+1)(x+1) = x^3+1$ (the middle terms cancel).

$(x-1)(x+1) = x^2-1$ .

…

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CBSE 2026Set ANNUAL1 mark
$\Delta = \begin{vmatrix}1 & 2 & 4\\ -1 & 3 & 0\\ 4 & 1 & 0\end{vmatrix}$
›Reveal solutionSolution

Expand the $3\times3$ determinant along the first row.

$\Delta=\begin{vmatrix}1&2&4\\-1&3&0\\4&1&0\end{vmatrix}$

Expanding along row 1: …

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CBSE 2026Set ANNUAL1 mark
$\Delta = \begin{vmatrix}0 & \sin\alpha & -\cos\alpha\\ -\sin\alpha & 0 & \sin\beta\\ \cos\alpha & -\sin\beta & 0\end{vmatrix}$
›Reveal solutionSolution

The matrix is skew-symmetric (each $a_{ij}=-a_{ji}$ ) and every odd-order skew-symmetric matrix has determinant $0$ .

Check: $a_{12}=\sin\alpha=-a_{21}$ , $a_{13}=-\cos\alpha=-a_{31}$ , $a_{23}=\sin\beta=-a_{32}$ , and all diagonal entries are $0$ — so the matrix is skew-symmetric.

…

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CBSE 2026Set ANNUAL1 markMCQ
$\begin{vmatrix} \cos 30^\circ & \sin 30^\circ \\ \sin 30^\circ & \cos 30^\circ \end{vmatrix} =$
(a) $\dfrac{1}{2}$
(b) $\dfrac{\sqrt{3}}{2}$
(c) 0
(d) None of these
›Reveal solutionSolution

This determinant has the form $\cos^2\theta - \sin^2\theta = \cos 2\theta$ .

$\begin{vmatrix} \cos 30^\circ & \sin 30^\circ \\ \sin 30^\circ & \cos 30^\circ \end{vmatrix} = \cos 30^\circ \cdot \cos 30^\circ - \sin 30^\circ \cdot \sin 30^\circ = \cos^2 30^\circ - \sin^2 30^\circ$

…

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CBSE 2026Set ANNUAL1 mark
$Q.Evaluate the determinant \Delta = \begin{vmatrix}1 & 2 & 4\ -1 & 3 & 0\ 4 & 1 & 0\end{vmatrix}.$
›Reveal solutionSolution

Expand the $3\times3$ determinant along the first row (or any row/column) using cofactors.

Working: Expanding along Row 1:

$\Delta = \begin{vmatrix}1 & 2 & 4\\ -1 & 3 & 0\\ 4 & 1 & 0\end{vmatrix}$ …

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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 65/4/11 markMCQ
$\det(M) = m$
›Reveal solutionSolution
The key idea is that $MN = mI$ implies $N = m M^{-1}$ , so $\det(N) = m^3 \det(M^{-1}) = m^3 \cdot \frac{1}{m} = m^2$ . The correct option is (D).

The problem gives us two square matrices $M$ and $N$ of order 3, with $\det(M) = m$ and $MN = mI$ , where $I$ is the $3 \times 3$ identity matrix. We need $\det(N)$ .

The central concept here is the relationship between matrix multiplication and determinants. When two matrices multiply to give a scalar times the identity, that scalar is intimately connected to the determinant of the first matrix. The equation $MN = mI$ is not just a product — it tells us that $N$ is essentially a scaled inverse of $M$ .

Why? Because if $MN = mI$ , then multiplying both sides on the left by $M^{-1}$ (assuming $M$ is invertible) gives $N = m M^{-1}$ . But we must first check: is $M$ invertible? Yes — since $\det(M) = m \neq 0$ (the problem doesn't state $m \neq 0$ explicitly, but if $m = 0$ , then $MN = 0$ , which would make $N$ singular and the answer ambiguous; in standard exam contexts, $m$ is taken as a non-zero scalar, often a real number, and the options suggest $m \neq 0$ ). So $M^{-1}$ exists.

Now, the determinant of a scalar multiple of a matrix: for an $n \times n$ matrix $A$ , $\det(kA) = k^n \det(A)$ . Here $n = 3$ , so $\det(m M^{-1}) = m^3 \det(M^{-1})$ .

And we know $\det(M^{-1}) = \frac{1}{\det(M)} = \frac{1}{m}$ .

Putting it together:

From $MN = mI$ , take determinant on both sides: $\det(MN) = \det(mI)$ .

$\det(MN) = \det(M) \cdot \det(N) = m \cdot \det(N)$ . …
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Q.(a) If $P = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}$ and $Q = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}$ , find $(QP)$ and hence solve the following system of equations using matrices: $x - y = 3,\ 2x + 3y + 4z = 17,\ y + 2z = 7$

Part (a)Concept understanding — Inverse Matrix Method

The Inverse Matrix Method

Writing the system as $AX=B$

The idea

When it works

Worked steps

Part (b)Concept understanding — Determinant Evaluation Using Identities

Determinant Evaluation Using Identities

The geometric intuition

The three row (or column) operations

The strategy

Worked example

Part (a) — $QP$ and the linear system

Part (a) — $QP$ and solving the system

Why determinants change under row operations

Step-by-step evaluation

Part (a) — QPQPQP and solving the system

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Part (a) — $QP$ and solving the system

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