CUET Mathematics 2024 - If P=[arrayr-1 \\ 2 \\ 1array] and Q=[arraylll2 & -4 & 1array] are two matrices, then (P Q)^ will be : | PYQs + Solutions

CUET Mathematics 2024

Algebra

Matrices & Determinants

Easy

If $P=\left[\begin{array}{r}-1 \\ 2 \\ 1\end{array}\right]$ and $Q=\left[\begin{array}{lll}2 & -4 & 1\end{array}\right]$ are two matrices, then $(P Q)^{\prime}$ will be :

\left[\begin{array}{ccc}4 & 5 & 7 \\ -3 & -3 & 0 \\ 0 & -3 & -2\end{array}\right]

\left[\begin{array}{rrr}-2 & 4 & 2 \\ 4 & -8 & -4 \\ -1 & 2 & 1\end{array}\right]

\left[\begin{array}{rrr}5 & 5 & 2 \\ 7 & 6 & 7 \\ -9 & -7 & 0\end{array}\right]

\left[\begin{array}{rrr}-2 & 4 & 8 \\ 7 & 5 & 7 \\ -8 & -2 & 6\end{array}\right]

✅ Correct Option: 2

Let's find

(PQ)'

where

P=\begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}

and

Q=\begin{bmatrix} 2 & -4 & 1 \end{bmatrix}

When multiplying a

3 \times 1

matrix with a

1 \times 3

matrix, we get a

3 \times 3

matrix.

PQ = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 2 & -4 & 1 \end{bmatrix}

= \begin{bmatrix} (-1)(2) & (-1)(-4) & (-1)(1) \\ (2)(2) & (2)(-4) & (2)(1) \\ (1)(2) & (1)(-4) & (1)(1) \end{bmatrix}

= \begin{bmatrix} -2 & 4 & -1 \\ 4 & -8 & 2 \\ 2 & -4 & 1 \end{bmatrix}

To find

(PQ)'

, we transpose the matrix by converting rows to columns:

(PQ)' = \begin{bmatrix} -2 & 4 & 2 \\ 4 & -8 & -4 \\ -1 & 2 & 1 \end{bmatrix}

Alternatively, we can use the property:

(PQ)' = Q'P'

P' = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}

Q' = \begin{bmatrix} 2 \\ -4 \\ 1 \end{bmatrix}

Q'P' = \begin{bmatrix} 2 \\ -4 \\ 1 \end{bmatrix} \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 4 & 2 \\ 4 & -8 & -4 \\ -1 & 2 & 1 \end{bmatrix}