(l) Find the product of $(2 \quad 3 \quad -1) \times \begin{pmatrix} 1 & 0 & 2 \\ 3 & -2 & 4 \\ 2 & 1 & 0 \end{pmatrix}$.

(l) Find the product of (2 quad 3 quad -1) × 1 0 2 \3 -2 4 \2 1 0 .

Question 20 of 37

(2 \quad 3 \quad -1) \times \begin{pmatrix} 1 & 0 & 2 \\ 3 & -2 & 4 \\ 2 & 1 & 0 \end{pmatrix}

ChseodishaCHSE Odisha Plus Two (Class 12) Commerce Board 2019Subjective· 2mImportance★★★★★est

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The product is the row matrix $(9\ \ -7\ \ 16)$ .

A $1 \times 3$ matrix times a $3 \times 3$ matrix gives a $1 \times 3$ matrix. Each entry is the row dotted with a column of the second matrix:

$(2\ \ 3\ \ -1)\begin{pmatrix} 1 & 0 & 2 \\ 3 & -2 & 4 \\ 2 & 1 & 0 \end{pmatrix}.$

1st entry (column 1): $2(1) + 3(3) + (-1)(2) = 2 + 9 - 2 = 9.$

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

CBSE 2022Set ANNUAL1 markMCQ
$\begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$
(a) $28$
(b) $[20]$
(c) $20$
(d) $[28]$
›Reveal solutionSolution

Inner product $2(1)+4(2)+6(3)=28$ , returned as a $1\times 1$ matrix $[28]$ .

The entries multiply pairwise and are added:

$2\times 1 + 4\times 2 + 6\times 3 = 2 + 8 + 18 = 28.$

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CBSE 2022Set ANNUAL1 markMCQ
$A = \begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix}$
(a) $\begin{bmatrix} -6 & -9 \\ 3 & -12 \end{bmatrix}$
(b) $\begin{bmatrix} -6 & -9 \\ 1 & 4 \end{bmatrix}$
(c) $\begin{bmatrix} 2 & 3 \\ -3 & -12 \end{bmatrix}$
(d) $\begin{bmatrix} 6 & 9 \\ 3 & 12 \end{bmatrix}$
›Reveal solutionSolution

$-3A$ scales each entry by $-3$ ⇒ option (a).

Scalar multiplication of a matrix multiplies every element by the scalar: …

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CBSE 2022Set ANNUAL1 markMCQ
$\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} + \begin{bmatrix} -1 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} + \begin{bmatrix} 9 & 8 & 7 \\ 6 & 5 & 4 \end{bmatrix}$
(a) $\begin{bmatrix} 9 & 8 & 7 \\ 6 & 5 & 4 \end{bmatrix}$
(b) $\begin{bmatrix} 9 & 8 & 11 \\ 6 & 5 & 12 \end{bmatrix}$
(c) $\begin{bmatrix} 11 & 12 & 11 \\ 14 & 15 & 12 \end{bmatrix}$
(d) $\begin{bmatrix} 0 & 0 & 7 \\ 0 & 0 & 4 \end{bmatrix}$
›Reveal solutionSolution

Element-wise addition ⇒ $\begin{bmatrix}9&8&11\\6&5&12\end{bmatrix}$ .

Matrices of the same order are added entry by entry:

$\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}+\begin{bmatrix}-1&-2&1\\-4&-5&2\end{bmatrix}+\begin{bmatrix}9&8&7\\6&5&4\end{bmatrix}.$

Row 1: $(1-1+9,\;2-2+8,\;3+1+7)=(9,\;8,\;11)$ . …

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Q.(l) Find the product of $(2 \quad 3 \quad -1) \times \begin{pmatrix} 1 & 0 & 2 \\ 3 & -2 & 4 \\ 2 & 1 & 0 \end{pmatrix}$ .

Concept understanding — Matrix Algebra: Addition, Subtraction and Multiplication

Q.(l) Find the product of (23−1)×(1023−24210)(2 \quad 3 \quad -1) \times \begin{pmatrix} 1 & 0 & 2 \\ 3 & -2 & 4 \\ 2 & 1 & 0 \end{pmatrix}(23−1)×​132​0−21​240​​.

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Q.(l) Find the product of $(2 \quad 3 \quad -1) \times \begin{pmatrix} 1 & 0 & 2 \\ 3 & -2 & 4 \\ 2 & 1 & 0 \end{pmatrix}$ .

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