Matrices: Solved Questions


Question 1 (ICSE 2018):

If\space A = \begin{bmatrix}2 & 3\\\ 5 & 7\end{bmatrix}, B = \begin{bmatrix}0 & 4\\\ -1 & 7\end{bmatrix} and \\ \space C = \begin{bmatrix}1 & 0\\\ -1 & 4\end{bmatrix}, find \space {AC+B^2-10C}.

Solution 1:

Step 1:
\\ AC=A \times C=\begin{bmatrix}2 & 3\\\ 5 & 7\end{bmatrix} \times \begin{bmatrix}1 & 0\\\ -1 & 4\end{bmatrix} \\ = \begin{bmatrix}(2 \times 1)+(3 \times -1) & (2 \times 0)+(3 \times 4)\\\ (5 \times 1)+(7 \times -1) & (5 \times 0)+(7 \times 4)\end{bmatrix} \\ \implies AC =\begin{bmatrix}(2-3) & (0+12)\\\ (5-7) & (0+28)\end{bmatrix} = \begin{bmatrix}-1 & 12\\\ -2 & 28\end{bmatrix}

Step 2:
\\ B^2 = B \times B=\begin{bmatrix}0 & 4\\\ -1 & 7\end{bmatrix} \times \begin{bmatrix}0 & 4\\\ -1 & 7\end{bmatrix} \\ = \begin{bmatrix}(0 \times 0)+(4 \times -1) & (0 \times 4)+(4 \times 7)\\\ (-1 \times 0)+(7 \times -1) & (-1 \times 4)+(7 \times 7)\end{bmatrix} \\ \implies B^2 =\begin{bmatrix}(0-4) & (0+28)\\\ (0-7) & (-4+49)\end{bmatrix} = \begin{bmatrix}-4 & 28\\\ -7 & 45\end{bmatrix}

Step 3:
\\ 10C = 10 \times C=10 \times {\begin{bmatrix}1 & 0\\\ -1 & 4\end{bmatrix}} = \begin{bmatrix}10\times 1 & 10 \times 0\\\ 10 \times -1 & 10 \times 4\end{bmatrix}\\ = \begin{bmatrix}10 & 0\\\ -10 & 40\end{bmatrix} \\ \implies 10C =\begin{bmatrix}10 & 0\\\ -10 & 40\end{bmatrix}

From Steps 1, 2 and 3 we have:
\\ AC+B^2-10C \\= \begin{bmatrix}-1 & 12\\\ -2 & 28\end{bmatrix} + \begin{bmatrix}-4 & 28\\\ -7 & 45\end{bmatrix} - \begin{bmatrix}10 & 0\\\ -10 & 40\end{bmatrix} \\ \implies AC+B^2-10C\\ =\begin{bmatrix}-1+(-4)-10 & 12+28-0\\\ -2+(-7)-(-10) & 28+45-40\end{bmatrix} \\ =\begin{bmatrix}-1-4-10 & 12+28-0\\\ -2-7+10 & 28+45-40\end{bmatrix}\\=\begin{bmatrix}-15 & 40\\\ 1 & 33\end{bmatrix}\space (Answer)

Question 2 (ICSE 2018):

Find the value of ‘x’ and ‘y’ if:
\\ 2\begin{bmatrix}x & 7\\\ 9 & y-5\end{bmatrix}+\begin{bmatrix}6 & -7\\\ 4 & 5\end{bmatrix} = \begin{bmatrix}10 & 7\\\ 22 & 15\end{bmatrix}

Solution 2:
\\ \begin{bmatrix}2\times x & 2\times 7\\\ 2\times9 & 2\times(y-5)\end{bmatrix} +\begin{bmatrix}6 & -7\\\ 4 & 5\end{bmatrix} = \begin{bmatrix}10 & 7\\\ 22 & 15\end{bmatrix} \\ \implies \begin{bmatrix}2x & 14\\\ 18 & 2y-10\end{bmatrix} +\begin{bmatrix}6 & -7\\\ 4 & 5\end{bmatrix} = \begin{bmatrix}10 & 7\\\ 22 & 15\end{bmatrix} \\ \implies \begin{bmatrix}2x+6 & 14+(-7)\\\ 18+4 & 2y-10+5\end{bmatrix} = \begin{bmatrix}10 & 7\\\ 22 & 15\end{bmatrix} \\ \implies \begin{bmatrix}2x+6 & 14-7\\\ 22 & 2y-5\end{bmatrix} = \begin{bmatrix}10 & 7\\\ 22 & 15\end{bmatrix}
We obtain two equations one each for x and y:
\\ 2x+6=10 \label{1} \\ 2y-5=15 \label{2} \\ \implies 2x=10-6 \implies 2x=4 \\\implies x=4/2 \\ and \ 2y-5=15 \implies 2y=15+5\implies 2y=20 \\\implies y=20/2 \\ \therefore x=2, y=10 \space(Answer)

Question 3:

\\ Simplify: \begin{bmatrix}-2sin \space30\textdegree & cosec \space30\textdegree\\\ tan 45\textdegree & cos0\textdegree\end{bmatrix} \\ \times\begin{bmatrix}cot 45\textdegree & sin 90\textdegree\\\ 2sec 0\textdegree & sec60\textdegree\end{bmatrix} \times \begin{bmatrix} cosec 90\textdegree \\\ 2cos 60\textdegree\end{bmatrix}

Solution 3:

Substituting values of trigonometrical functions:

\\ \begin{bmatrix}-2\times \frac{1}{2} & 2\\\ 1 & 1\end{bmatrix}\times\begin{bmatrix}1 & 1\\\ 2 & 2\end{bmatrix}\times \begin{bmatrix} 1 \\\ 2\times \frac{1}{2}\end{bmatrix}\\ \implies \begin{bmatrix}-1 & 2\\\ 1 & 1\end{bmatrix}\times\begin{bmatrix}1 & 1\\\ 2 & 2\end{bmatrix}\times \begin{bmatrix} 1 \\\ 1\end{bmatrix}\\ \implies \begin{bmatrix}(-1\times 1+2\times 2) & (-1\times 1+2\times 2)\\\ (1\times1+1\times2) & (1\times1+1\times2)\end{bmatrix}\times \begin{bmatrix} 1 \\\ 1\end{bmatrix}\\ \implies \begin{bmatrix}3 & 3\\\ 3 & 3\end{bmatrix}\times \begin{bmatrix} 1 \\\ 1\end{bmatrix}\\ \implies \begin{bmatrix} (3\times1+3\times1)\\\ (3\times1+ 3\times1)\end{bmatrix} = \begin{bmatrix}6\\\ 6\end{bmatrix}\space (Answer)

Question 4:

\text {Given Matrix B}=\begin{bmatrix}1 & 1\\\ 8&3\end{bmatrix}.\\ \text{Find Matrix X, if } X=B^2-4B. \\ \text{Hence, solve for 'a' and 'b' \\given X}\begin{bmatrix}a\\\ b\end{bmatrix}=\begin{bmatrix}5\\\ 50\end{bmatrix}\\

Solution 4:

\\ B^2=\begin{bmatrix}1 & 1\\\ 8&3\end{bmatrix} \times \begin{bmatrix}1 & 1\\\ 8&3\end{bmatrix}\\ =\begin{bmatrix}(1\times1+1\times8) & (1\times1+1\times3)\\\ (8\times1+3\times8) &(8\times1+3\times3)\end{bmatrix} \\ =\begin{bmatrix}(1+8) & (1+3)\\\ (8+24) &(8+9)\end{bmatrix} \\ \therefore B^2=\begin{bmatrix}9 & 4\\\ 32 &17\end{bmatrix} \\ Now, X=B^2-4B=\begin{bmatrix}9 & 4\\\ 32 &17\end{bmatrix}-4\times\begin{bmatrix}1 & 1\\\ 8&3\end{bmatrix}\\ =\begin{bmatrix}9 & 4\\\ 32 &17\end{bmatrix}-\begin{bmatrix}4\times1 & 4\times1\\\ 4\times8&4\times3\end{bmatrix}\\ =\begin{bmatrix}9 & 4\\\ 32 &17\end{bmatrix}-\begin{bmatrix}4 & 4\\\ 32&12\end{bmatrix}\\ =\begin{bmatrix}(9-4) & (4-4)\\\ (32-32) &(17-12)\end{bmatrix}\\ \therefore X=\begin{bmatrix}5& 0\\\ 0&5\end{bmatrix}\\ Hence, \begin{bmatrix}5& 0\\\ 0&5\end{bmatrix}\begin{bmatrix}a\\\ b\end{bmatrix}=\begin{bmatrix}5\\\ 50\end{bmatrix}\\ \implies \begin{bmatrix}5& 0\\\ 0&5\end{bmatrix}\begin{bmatrix}a\\\ b\end{bmatrix}=\begin{bmatrix}5\\\ 50\end{bmatrix}\\ \implies \begin{bmatrix}(5\times a+0\times b)\\\ (0\times a+5\times b)\end{bmatrix} =\begin{bmatrix}5\\\ 50\end{bmatrix}\\ \implies \begin{bmatrix}5a\\\ 5b\end{bmatrix} =\begin{bmatrix}5\\\ 50\end{bmatrix}\\ \implies 5\times \begin{bmatrix}a\\\ b\end{bmatrix} =5\times\begin{bmatrix}1\\\ 10\end{bmatrix}\\ \implies \begin{bmatrix}a\\\ b\end{bmatrix} =\begin{bmatrix}1\\\ 10\end{bmatrix}\\ \therefore a=1,b=10 (Answer)

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