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Matrices | Matrices Exercise 8.2

Question 5

\text { If } A=\left[\begin{array}{cc} 0 & -1 \\ 1 & 2 \end{array}\right] \text { and } B=\left[\begin{array}{cc} 1 & 2 \\ -1 & 1 \end{array}\right]

Find the matrix X if:

(i) 3A + X = B

(ii) X – 3B = 2A

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Given

\begin{array}{l} A=\left[\begin{array}{cc} 0 & -1 \\ 1 & 2 \end{array}\right] \\ B=\left[\begin{array}{cc} 1 & 2 \\ -1 & 1 \end{array}\right] \end{array}

Now, we have to find

\begin{aligned} &\text { (i) } 3 A+X=B\\ &X=B-3 A \end{aligned}

\begin{aligned} &\text { Substituting the values we get }\\ &\begin{aligned} X &=\left[\begin{array}{rr} 1 & 2 \\ -1 & 1 \end{array}\right]-3\left[\begin{array}{rr} 0 & -1 \\ 1 & 2 \end{array}\right] \\ &=\left[\begin{array}{rr} 1 & 2 \\ -1 & 1 \end{array}\right]-\left[\begin{array}{rr} 0 & -3 \\ 3 & 6 \end{array}\right] \\ &=\left[\begin{array}{rr} 1-0 & 2+3 \\ -1-3 & 1-6 \end{array}\right]=\left[\begin{array}{rr} 1 & 5 \\ -4 & -5 \end{array}\right] \end{aligned} \end{aligned}

\begin{aligned} &\text { (ii) } X-3 B=2 A\\ &X=2 A+3 B \end{aligned}

\begin{aligned} &\text { Now substituting the values } A \text { and } B \text { we get }\\ &X=2\left[\begin{array}{rr} 0 & -1 \\ 1 & 2 \end{array}\right]+3\left[\begin{array}{rr} 1 & 2 \\ -1 & 1 \end{array}\right]\\ &=\left[\begin{array}{rr} 0 & -2 \\ 2 & 4 \end{array}\right]+\left[\begin{array}{rr} 3 & 6 \\ -3 & 3 \end{array}\right]\\ &=\left[\begin{array}{rr} 0+3 & -2+6 \\ 2-3 & 4+3 \end{array}\right]=\left[\begin{array}{rr} 3 & 4 \\ -1 & 7 \end{array}\right] \end{aligned}

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Question 5

\text { If } A=\left[\begin{array}{cc} 0 & -1 \\ 1 & 2 \end{array}\right] \text { and } B=\left[\begin{array}{cc} 1 & 2 \\ -1 & 1 \end{array}\right]

Find the matrix X if:

(i) 3A + X = B

(ii) X – 3B = 2A

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

Given

\begin{array}{l} A=\left[\begin{array}{cc} 0 & -1 \\ 1 & 2 \end{array}\right] \\ B=\left[\begin{array}{cc} 1 & 2 \\ -1 & 1 \end{array}\right] \end{array}

Now, we have to find

\begin{aligned} &\text { (i) } 3 A+X=B\\ &X=B-3 A \end{aligned}

\begin{aligned} &\text { Substituting the values we get }\\ &\begin{aligned} X &=\left[\begin{array}{rr} 1 & 2 \\ -1 & 1 \end{array}\right]-3\left[\begin{array}{rr} 0 & -1 \\ 1 & 2 \end{array}\right] \\ &=\left[\begin{array}{rr} 1 & 2 \\ -1 & 1 \end{array}\right]-\left[\begin{array}{rr} 0 & -3 \\ 3 & 6 \end{array}\right] \\ &=\left[\begin{array}{rr} 1-0 & 2+3 \\ -1-3 & 1-6 \end{array}\right]=\left[\begin{array}{rr} 1 & 5 \\ -4 & -5 \end{array}\right] \end{aligned} \end{aligned}

\begin{aligned} &\text { (ii) } X-3 B=2 A\\ &X=2 A+3 B \end{aligned}

\begin{aligned} &\text { Now substituting the values } A \text { and } B \text { we get }\\ &X=2\left[\begin{array}{rr} 0 & -1 \\ 1 & 2 \end{array}\right]+3\left[\begin{array}{rr} 1 & 2 \\ -1 & 1 \end{array}\right]\\ &=\left[\begin{array}{rr} 0 & -2 \\ 2 & 4 \end{array}\right]+\left[\begin{array}{rr} 3 & 6 \\ -3 & 3 \end{array}\right]\\ &=\left[\begin{array}{rr} 0+3 & -2+6 \\ 2-3 & 4+3 \end{array}\right]=\left[\begin{array}{rr} 3 & 4 \\ -1 & 7 \end{array}\right] \end{aligned}

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