Question 12 Marks
For what values of $x : \left[\begin{array}{lll}{1} & {2} & {1}\end{array}\right] \left[\begin{array}{lll}{1} & {2} & {0} \\ {2} & {0} & {1} \\ {1} & {0} & {2}\end{array}\right]\left[\begin{array}{l}{0} \\ {2} \\ {x}\end{array}\right] = 0.$
Answer$\left[\begin{array}{lll}{1} & {2} & {1}\end{array}\right] \left[\begin{array}{lll}{1} & {2} & {0} \\ {2} & {0} & {1} \\ {1} & {0} & {2}\end{array}\right] \left[\begin{array}{l}{0} \\ {2} \\ {x}\end{array}\right] = 0$
$\Rightarrow \left[\begin{array}{lll}{1+4+1} & {2+0+0} & {0+2+2}\end{array}\right] \left[\begin{array}{l}{0} \\ {2} \\ {x}\end{array}\right] = 0$
$\Rightarrow \left[\begin{array}{lll}{6} & {2} & {4}\end{array}\right] \left[\begin{array}{l}{0} \\ {2} \\ {x}\end{array}\right] = 0$
$\Rightarrow [6(0) + 2(2) + 4(x)] = 0$
$\Rightarrow [4 + 4x] = [0]$
$\therefore 4 + 4x = 0$
$\Rightarrow x = -1$.
Therefore,the required value of $x$ is $-1.$
View full question & answer→Question 22 Marks
Show that the matrix B’AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.
Answer(B'AB)' = [B'(AB)]' = (AB)' (B')' [$\because$ (CD)' = D'C']
$\Rightarrow$ (B'AB)' = B'A'B ……….(i)
Case I: A is a symmetric matrix,
$\Rightarrow$ A' = A
$\therefore$ From eq. (i) (B'AB)' = B'AB
$\therefore$ B'AB is a symmetric matrix.
Case II: A is a skew symmetric matrix.
$\Rightarrow$ A' = - A
Putting A' = - A in eq. (i),
(B'AB)' = B'(- A)B = - B'AB
$\therefore$ B'AB is a skew symmetric matrix.
View full question & answer→Question 32 Marks
If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.
AnswerA and B are symmetric matrices. $\Rightarrow$ A’ = A and B’ = B ……….(i)
Now, (AB - BA)’ = (AB)’ - (BA)’ $\Rightarrow$ (AB - BA)’ = B’A’ - A’B’ [Reversal law]
$\Rightarrow$ (AB - BA)’ = BA - AB [Using eq. (i)]
$\Rightarrow$ (AB - BA)’ = - (AB - BA)
Therefore, (AB - BA) is a skew symmetric.
View full question & answer→Question 42 Marks
For the matrix $A=\left[\begin{array}{ll} {1} & {5} \\ {6} & {7} \end{array}\right]$, verify that (A + A′) is a symmetric matrix.
AnswerHere,
$A=\left[\begin{array}{ll} {1} & {5} \\ {6} & {7} \end{array}\right] \Rightarrow A^{\prime}=\left[\begin{array}{ll} {1} & {6} \\ {5} & {7} \end{array}\right]$
On adding them we get,
$A+A^{\prime}=\left[\begin{array}{ll} {1} & {5} \\ {6} & {7} \end{array}\right]+\left[\begin{array}{ll} {1} & {6} \\ {5} & {7} \end{array}\right]$
$\Rightarrow A+A^{\prime}=\left[\begin{array}{cc} {1+1} & {5+6} \\ {6+5} & {7+7} \end{array}\right]$
$\Rightarrow \mathrm{A}+\mathrm{A}^{\prime}=\left[\begin{array}{ll} {2} & {11} \\ {11} & {14} \end{array}\right] $ ...(1)
Now, $(\mathrm{A}+\mathrm{A}^{\prime})^\prime=A^\prime+A=$ $\left(\mathrm{A}+\mathrm{A}^{\prime}\right)=\left[\begin{array}{ll} {2} & {11} \\ {11} & {14} \end{array}\right] $ ...(2)
So, from equation (1) & (2) we get,
(A + A') = (A + A')', hence we can say that (A + A') is a Symmetric matrix.
View full question & answer→Question 52 Marks
If $A=\left[\begin{array}{cc} {\sin \alpha} & {\cos \alpha} \\ {-\cos \alpha} & {\sin \alpha} \end{array}\right]$, then verify that A′ A = I
AnswerHere, $A=\left[\begin{array}{cc} {\sin \alpha} & {\cos \alpha} \\ {-\cos \alpha} & {\sin \alpha} \end{array}\right]$
$\therefore$ $A^{\prime}=\left[\begin{array}{cc} {\sin \alpha} & {-\cos \alpha} \\ {\cos \alpha} & {\sin \alpha} \end{array}\right]$
$\Rightarrow$ $A A^{\prime}=\left[\begin{array}{cc} {\sin \alpha} & {\cos \alpha} \\ {-\cos \alpha} & {\sin \alpha} \end{array}\right] \times\left[\begin{array}{cc} {\sin \alpha} & {-\cos \alpha} \\ {\cos \alpha} & {\sin \alpha} \end{array}\right]$
$\Rightarrow~ \mathrm{AA}^{\prime}=\left[\begin{array}{ll} {\sin \alpha \times \sin \alpha+\cos \alpha \times \cos \alpha} & {\sin \alpha \times(-\cos \alpha)+\cos \alpha \times \sin \alpha} \\ {-\cos \alpha \times \sin \alpha+\sin \alpha \times \cos \alpha} & {-\cos \alpha \times(-\cos \alpha)+\sin \alpha \times \sin \alpha} \end{array}\right]$
$\Rightarrow \mathrm{AA}^{\prime}=\left[\begin{array}{cc} {\sin ^{2} \alpha+\cos ^{2} \alpha} & {-\sin \alpha \cos \alpha+\cos \alpha \sin \alpha} \\ {-\cos \alpha \sin \alpha+\sin \alpha \cos \alpha} & {\cos ^{2} \alpha+\sin ^{2} \alpha} \end{array}\right]$
$\Rightarrow \mathrm{AA}^{\prime}=\left[\begin{array}{ll} {1} & {0} \\ {0} & {1} \end{array}\right] = I \quad\left(\because \cos ^{2} \alpha+\sin ^{2} \alpha=1\right)$
Therefore, AA' = I
Hence verified.
View full question & answer→Question 62 Marks
If $A=\left[\begin{array}{cc} {\cos \alpha} & {\sin \alpha} \\ {-\sin \alpha} & {\cos \alpha} \end{array}\right]$, then verify that A′ A = I
AnswerWe know A’ can be calculated by taking the transpose of the given matrix A.
Therefore, $A^{\prime}=\left[\begin{array}{ll} {\cos \alpha} & {-\sin \alpha} \\ {\sin \alpha} & {\cos \alpha} \end{array}\right]$
Now multiply A and A’. So,
$A A^{\prime}=\left[\begin{array}{cc} {\cos \alpha} & {\sin \alpha} \\ {-\sin \alpha} & {\cos \alpha} \end{array}\right] \times\left[\begin{array}{cc} {\cos \alpha} & {-\sin \alpha} \\ {\sin \alpha} & {\cos \alpha} \end{array}\right]$
$\Rightarrow \mathrm{AA}^{\prime}=\left[\begin{array}{cc} {(\cos \alpha \times \cos \alpha)+(\sin \alpha) \times(\sin \alpha)} & {\cos \alpha \times(-\sin \alpha)+(\sin \alpha) \times \cos \alpha} \\ {-\sin \alpha \times \cos \alpha+\cos \alpha \times(\sin \alpha)} & {-\sin \alpha \times(-\sin \alpha)+\cos \alpha \times \cos \alpha} \end{array}\right]$
$\Rightarrow \mathrm{AA}^{\prime}=\left[\begin{array}{cc} {\cos ^{2} \alpha+\sin ^{2} \alpha} & {-\sin \alpha \cos \alpha+\cos \alpha \sin \alpha} \\ {-\sin \alpha \cos \alpha+\cos \alpha \sin \alpha} & {\sin ^{2} \alpha+\cos ^{2} \alpha} \end{array}\right]$
$\Rightarrow \mathrm{AA}^{\prime}=\left[\begin{array}{ll} {1} & {0} \\ {0} & {1} \end{array}\right] ...(1) \quad\left(\because \cos \alpha^{2}+\sin \alpha^{2}=1\right)$
And we know ‘I’ represents an identity matrix
Therefore, $I=\left[\begin{array}{ll} {1} & {0} \\ {0} & {1} \end{array}\right] $...(2)
From equation 1 & 2 we can say that
AA’ = I
AA’ = I. Hence verified.
View full question & answer→Question 72 Marks
For the matrices, $A$ and $B,$ verify that $(AB)\ ' = B\ 'A\ ',$ where $A=\left[\begin{array}{l} {0} \\ {1} \\ {2} \end{array}\right], B=\left[\begin{array}{lll} {1} & {5} & {7} \end{array}\right]$
AnswerGiven that, $A=\left[\begin{array}{l} {0} \\ {1} \\ {2} \end{array}\right] \text { and } B=\left[\begin{array}{lll} {1} & {5} & {7} \end{array}\right]$
Explanation: The product of two matrices is defined or possible only if the number of columns of the former matrix is equal to number of rows of the latter matrix.
$A B=\left[\begin{array}{l} {0} \\ {1} \\ {2} \end{array}\right] \left[\begin{array}{lll} {1} & {5} & {7} \end{array}\right]$
$\Rightarrow \mathrm{AB}=\left[\begin{array}{ccc} {0 \times 1} & {0 \times 5} & {0 \times 7} \\ {1 \times 1} & {1 \times 5} & {1 \times 7} \\ {2 \times 1} & {2 \times 5} & {2 \times 7} \end{array}\right]$
$\Rightarrow \mathrm{AB}=\left[\begin{array}{ccc} {0} & {0} & {0} \\ {1} & {5} & {7} \\ {2} & {10} & {14} \end{array}\right]$
Therefore, $(\mathrm{AB})^\prime=\left[\begin{array}{ccc} {0} & {1} & {2} \\ {0} & {5} & {10} \\ {0} & {7} & {14} \end{array}\right] ........(1)$
Now, $A^{\prime}=\left[\begin{array}{lll} {0} & {1} & {2} \end{array}\right] \text { and } B^{\prime}=\left[\begin{array}{l} {1} \\ {5} \\ {7} \end{array}\right]$
Therefore, $\mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{l} {1} \\ {5} \\ {7} \end{array}\right] \left[\begin{array}{lll} {0} & {1} & {2} \end{array}\right]$
$\Rightarrow \mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{ccc} {1 \times 0} & {1 \times 1} & {1 \times 2} \\ {5 \times 0} & {5 \times 1} & {5 \times 2} \\ {7 \times 0} & {7 \times 1} & {7 \times 2} \end{array}\right]$
$\Rightarrow \mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{ccc} {0} & {1} & {2} \\ {0} & {5} & {10} \\ {0} & {7} & {14} \end{array}\right] ......(2)$
From equation $(1) \ (2)$ we see that
$(AB)\ ' = B\ 'A\ '$. Hence verified.
View full question & answer→Question 82 Marks
For the matrices, A and B, verify that (AB)′ = B′A′, where $A=\left[\begin{array}{c} {1} \\ {-4} \\ {3} \end{array}\right], B=\left[\begin{array}{ccc} {-1} & {2} & {1} \end{array}\right]$
Answer$A=\left[\begin{array}{c} {1} \\ {-4} \\ {3} \end{array}\right] \text { and } B=\left[\begin{array}{ccc} {-1} & {2} & {1} \end{array}\right]$
Explanation: The product of two matrices is defined or possible only if the number of columns of the former matrix is equal to a number of rows of the latter matrix.
$A B=\left[\begin{array}{c} {1} \\ {-4} \\ {3} \end{array}\right] \left[\begin{array}{ccc} {-1} & {2} & {1} \end{array}\right]$
$\Rightarrow \mathrm{AB}=\left[\begin{array}{ccc} {1 \times(-1)} & {1 \times 2} & {1 \times 1} \\ {-4 \times(-1)} & {-4 \times 2} & {-4 \times 1} \\ {3 \times(-1)} & {3 \times 2} & {3 \times 1} \end{array}\right]$
$\Rightarrow A B=\left[\begin{array}{ccc} {-1} & {2} & {1} \\ {4} & {-8} & {-4} \\ {-3} & {6} & {3} \end{array}\right]$
So, $(\mathrm{AB})^\prime=\left[\begin{array}{ccc} {-1} & {4} & {-3} \\ {2} & {-8} & {6} \\ {1} & {-4} & {3} \end{array}\right] ~~~~~~...(i)$
$\mathrm{A}^{\prime}=\left[\begin{array}{lll} {1} & {-4} & {3} \end{array}\right] \text { and } \mathrm{B}^{\prime}=\left[\begin{array}{c} {-1} \\ {2} \\ {1} \end{array}\right]$
Therefore, $\mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{c} {-1} \\ {2} \\ {1} \end{array}\right] \times\left[\begin{array}{ccc} {1} & {-4} & {3} \end{array}\right]$
$\Rightarrow \mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{ccc} {-1 \times 1} & {-1 \times(-4)} & {-1 \times 3} \\ {2 \times 1} & {2 \times(-4)} & {2 \times 3} \\ {1 \times 1} & {1 \times(-4)} & {1 \times 3} \end{array}\right]$
$\Rightarrow \mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{ccc} {-1} & {4} & {-3} \\ {2} & {-8} & {6} \\ {1} & {-4} & {3} \end{array}\right] ....(ii)$
From equation (i) & (ii) we see that
(AB)’ = B’A’. Hence verified.
View full question & answer→Question 92 Marks
If A' = $\left[\begin{array}{cc} {-2} & {3} \\ {1} & {2} \end{array}\right] \text { and } \mathbf{B}=\left[\begin{array}{rr} {-1} & {0} \\ {1} & {2} \end{array}\right]$, then find (A + 2B)′
AnswerHere,
$A^\prime$ = $\left[\begin{array}{cc} {-2} & {1} \\ {3} & {2} \end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{cc} {-1} & {0} \\ {1} & {2} \end{array}\right] $
$\Rightarrow A=\left[\begin{array}{cc} {-2} & {3} \\ {1} & {2} \end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{cc} {-1} & {0} \\ {1} & {2} \end{array}\right]$
So, $A+2 B=\left[\begin{array}{rr} {-2} & {1} \\ {3} & {2} \end{array}\right]+2\left[\begin{array}{cc} {-1} & {0} \\ {1} & {2} \end{array}\right]$
$\Rightarrow A+2 B=\left[\begin{array}{cc} {-2} & {1} \\ {3} & {2} \end{array}\right]+\left[\begin{array}{cc} {-2} & {0} \\ {2} & {4} \end{array}\right]$
$\Rightarrow A+2 B=\left[\begin{array}{rr} {-4} & {1} \\ {5} & {6} \end{array}\right]$
Now, (A+2B)' = transpose of A+2B
$\Rightarrow(A+2 B)^\prime=\left[\begin{array}{cc} {-4} & {5} \\ {1} & {6} \end{array}\right]$
View full question & answer→Question 102 Marks
If $A^{\prime}=\left[\begin{array}{cc} {3} & {4} \\ {-1} & {2} \\ {0} & {1} \end{array}\right] \text { and } B=\left[\begin{array}{rrr} {-1} & {2} & {1} \\ {1} & {2} & {3} \end{array}\right]$ then verify (A – B)′ = A′ – B′
AnswerGiven, $A^{\prime}=\left[\begin{array}{cc} {3} & {4} \\ {-1} & {2} \\ {0} & {1} \end{array}\right] \text { and } B=\left[\begin{array}{rrr} {-1} & {2} & {1} \\ {1} & {2} & {3} \end{array}\right]$
Now, A = $\left[\begin{array}{rrr} {3} & {-1} & {0} \\ {4} & {2} & {1} \end{array}\right]$
$\Rightarrow A-B=\left[\begin{array}{ccc} {3} & {-1} & {0} \\ {4} & {2} & {1} \end{array}\right]-\left[\begin{array}{ccc} {-1} & {2} & {1} \\ {1} & {2} & {3} \end{array}\right]$
$\Rightarrow A-B=\left[\begin{array}{ccc} {3-(-1)} & {-1-2} & {0-1} \\ {4-1} & {2-2} & {1-3} \end{array}\right]$
$\Rightarrow A-B=\left[\begin{array}{ccc} {4} & {-3} & {-1} \\ {3} & {0} & {-2} \end{array}\right]$
Therefore, $(A-B)^\prime$ = $ \left[\begin{array}{cc} {4} & {3} \\ {-3} & {0} \\ {-1} & {-2} \end{array}\right] $
$~~~~~~A^{\prime}-B^{\prime}=\left[\begin{array}{cc} {3} & {4} \\ {-1} & {2} \\ {0} & {1} \end{array}\right]-\left[\begin{array}{cc} {-1} & {1} \\ {2} & {2} \\ {1} & {3} \end{array}\right]$
$\Rightarrow \mathrm{A}^{\prime}-\mathrm{B}^{\prime}=\left[\begin{array}{cc} {4} & {3} \\ {-3} & {0} \\ {-1} & {-2} \end{array}\right] $
From equation (1) & (2) we see that
(A - B)’ = A’ - B’. Hence verified.
View full question & answer→Question 112 Marks
If $A^{\prime}=\left[\begin{array}{cc} {3} & {4} \\ {-1} & {2} \\ {0} & {1} \end{array}\right] \text { and } B=\left[\begin{array}{rrr} {-1} & {2} & {1} \\ {1} & {2} & {3} \end{array}\right]$ then verify (A + B)′ = A′ + B′
AnswerGiven, $A^{\prime}=\left[\begin{array}{cc} {3} & {4} \\ {-1} & {2} \\ {0} & {1} \end{array}\right] \text { and } B=\left[\begin{array}{rrr} {-1} & {2} & {1} \\ {1} & {2} & {3} \end{array}\right]$
$\Rightarrow A=\left[\begin{array}{ccc} {3} & {-1} & {0} \\ {4} & {2} & {1} \end{array}\right]$
Now, A + B = $\left[\begin{array}{ccc} {3} & {-1} & {0} \\ {4} & {2} & {1} \end{array}\right]+\left[\begin{array}{ccc} {-1} & {2} & {1} \\ {1} & {2} & {3} \end{array}\right]$
$\Rightarrow A+B=\left[\begin{array}{ccc} {3+(-1)} & {-1+2} & {0+1} \\ {4+1} & {2+2} & {1+3} \end{array}\right]$
$\Rightarrow A+B=\left[\begin{array}{lll} {2} & {1} & {1} \\ {5} & {4} & {4} \end{array}\right]$
$\Rightarrow (A+B)' =\left[\begin{array}{ll} {2} & {5} \\ {1} & {4} \\ {1} & {4} \end{array}\right]$
Therefore, $(A+B)^\prime$ = $\left[\begin{array}{ll} {2} & {5} \\ {1} & {4} \\ {1} & {4} \end{array}\right]$...(1)
Now, A' + B' = $\left[\begin{array}{cc} {3} & {4} \\ {-1} & {2} \\ {0} & {1} \end{array}\right]+\left[\begin{array}{cc} {-1} & {1} \\ {2} & {2} \\ {1} & {3} \end{array}\right]$
$\Rightarrow A^{\prime}+B^{\prime}=\left[\begin{array}{ll} {2} & {5} \\ {1} & {4} \\ {1} & {4} \end{array}\right]$...(2)
From equation (1) & (2) we verify that
(A+B)’ = A’+B’. Hence verified.
View full question & answer→Question 122 Marks
If A = $\left[\begin{array}{ccc} {-1} & {2} & {3} \\ {5} & {7} & {9} \\ {-2} & {1} & {1} \end{array}\right]$ and B = $\left[\begin{array}{rrr} {-4} & {1} & {-5} \\ {1} & {2} & {0} \\ {1} & {3} & {1} \end{array}\right]$, then verify (A – B)′ = A′ – B′
AnswerWe will first calculate L.H.S i.e. (A+B)’ and then consecutively we will calculate R.H.S and verify that both are equal.
So, A - B = $\left[\begin{array}{ccc} {-1} & {2} & {3} \\ {5} & {7} & {9} \\ {-2} & {1} & {1} \end{array}\right]-\left[\begin{array}{ccc} {-4} & {1} & {-5} \\ {1} & {2} & {0} \\ {1} & {3} & {1} \end{array}\right]$
$\Rightarrow A-B=\left[\begin{array}{ccc} {-1-(-4)} & {2-1} & {3-(-5)} \\ {5-1} & {7-2} & {9-0} \\ {-2-1} & {1-3} & {1-1} \end{array}\right]$
$\Rightarrow A-B=\left[\begin{array}{ccc} {3} & {1} & {8} \\ {4} & {5} & {9} \\ {-3} & {-2} & {0} \end{array}\right]$
Therefore, $(A-B)^\prime$ = $\left[\begin{array}{ccc} {3} & {4} & {-3} \\ {1} & {5} & {-2} \\ {8} & {9} & {0} \end{array}\right]$ ...(1)
Now, $A^{\prime}=\left[\begin{array}{ccc} {-1} & {5} & {-2} \\ {2} & {7} & {1} \\ {3} & {9} & {1} \end{array}\right] \text { and } B^{\prime}=\left[\begin{array}{ccc} {-4} & {1} & {1} \\ {1} & {2} & {3} \\ {-5} & {0} & {1} \end{array}\right]$
So, A' - B' = $\left[\begin{array}{ccc} {-1} & {5} & {-2} \\ {2} & {7} & {1} \\ {3} & {9} & {1} \end{array}\right]-\left[\begin{array}{ccc} {-4} & {1} & {1} \\ {1} & {2} & {3} \\ {-5} & {0} & {1} \end{array}\right]$
$\Rightarrow A^{\prime}-B^{\prime}\left[\begin{array}{ccc} {-1-(-4)} & {5-1} & {-2-1} \\ {2-1} & {7-2} & {1-3} \\ {3-(-5)} & {9-0} & {1-1} \end{array}\right]$
$\Rightarrow \mathrm{A}^{\prime}-\mathrm{B}^{\prime}=\left[\begin{array}{ccc} {3} & {4} & {-3} \\ {1} & {5} & {-2} \\ {8} & {9} & {0} \end{array}\right] $...(2)
From equation (1) & (2) we verify that
$(A-B)^\prime=A^\prime-B^\prime$.
Hence verified.
View full question & answer→Question 132 Marks
If A = $\left[\begin{array}{ccc} {-1} & {2} & {3} \\ {5} & {7} & {9} \\ {-2} & {1} & {1} \end{array}\right]$ and B = $\left[\begin{array}{rrr} {-4} & {1} & {-5} \\ {1} & {2} & {0} \\ {1} & {3} & {1} \end{array}\right]$, then verify (A + B)′ = A′ + B′,
AnswerA = $\left[\begin{array}{ccc} {-1} & {2} & {3} \\ {5} & {7} & {9} \\ {-2} & {1} & {1} \end{array}\right] \text { and } B=\left[\begin{array}{ccc} {-4} & {1} & {-5} \\ {1} & {2} & {0} \\ {1} & {3} & {1} \end{array}\right]$
(A+B)’ = A’+B’
Explanation: We will first calculate L.H.S i.e. (A+B)’ and then consecutively we will calculate R.H.S and verify that both are equal.
So, A + B = $\left[\begin{array}{ccc} {-1} & {2} & {3} \\ {5} & {7} & {9} \\ {-2} & {1} & {1} \end{array}\right]+\left[\begin{array}{ccc} {-4} & {1} & {-5} \\ {1} & {2} & {0} \\ {1} & {3} & {1} \end{array}\right]$
$\Rightarrow A+B=\left[\begin{array}{ccc} {-1+(-4)} & {2+1} & {3+(-5)} \\ {5+1} & {7+2} & {9+0} \\ {-2+1} & {1+3} & {1+1} \end{array}\right]$
$\Rightarrow A+B=\left[\begin{array}{ccc} {-5} & {3} & {-2} \\ {6} & {9} & {9} \\ {-1} & {4} & {2} \end{array}\right]$
Therefore, $(A+B)^\prime=\left[\begin{array}{ccc} {-5} & {6} & {-1} \\ {3} & {9} & {4} \\ {-2} & {9} & {2} \end{array}\right] $ ...(1)
Noe, $A^{\prime}=\left[\begin{array}{ccc} {-1} & {5} & {-2} \\ {2} & {7} & {1} \\ {3} & {9} & {1} \end{array}\right] \text { and } B^{\prime}=\left[\begin{array}{ccc} {-4} & {1} & {1} \\ {1} & {2} & {3} \\ {-5} & {0} & {1} \end{array}\right]$
So, A' + B' = $\left[\begin{array}{ccc} {-1} & {5} & {-2} \\ {2} & {7} & {1} \\ {3} & {9} & {1} \end{array}\right]+\left[\begin{array}{ccc} {-4} & {1} & {1} \\ {1} & {2} & {3} \\ {-5} & {0} & {1} \end{array}\right]$
$\Rightarrow \mathrm{A}^{\prime}+\mathrm{B}^{\prime}=\left[\begin{array}{ccc} {-1+(-4)} & {5+1} & {-2+1} \\ {2+1} & {7+2} & {1+3} \\ {3+(-5)} & {9+0} & {1+1} \end{array}\right]$
$\Rightarrow A^{\prime}+B^{\prime}=\left[\begin{array}{ccc} {-5} & {6} & {-1} \\ {3} & {9} & {4} \\ {-2} & {9} & {2} \end{array}\right] $ ...(2)
From equation (1) &( 2), we see that
(A+B)’ = A’+B’. Hence verified.
View full question & answer→Question 142 Marks
Express matrix as the sum of a symmetric and a skew-symmetric matrix: $\left[\begin{array}{rr} {1} & {5} \\ {-1} & {2} \end{array}\right]$
AnswerGiven A =$\left[\begin{array}{cc} {1} & {5} \\ {-1} & {2} \end{array}\right]$
Explanation: As per Theorem “Any square matrix can be expressed as the sum of a symmetric and skew-symmetric matrix.” So in order to prove this we will be using Theorem which states that “For any square matrix A with real number entries, A + A’ is a symmetric matrix and A – A’ is a skew-symmetric matrix.”
Now, Let A = $\left[\begin{array}{cc} {1} & {5} \\ {-1} & {2} \end{array}\right]$
Therefore, $A^{\prime}=\left[\begin{array}{cc} {1} & {-1} \\ {5} & {2} \end{array}\right]$
Now, on adding A and A’ we will get,
$A+A^{\prime}=\left[\begin{array}{cc} {1} & {5} \\ {-1} & {2} \end{array}\right]+\left[\begin{array}{cc} {1} & {-1} \\ {5} & {2} \end{array}\right]$
$\Rightarrow A+A^{\prime}=\left[\begin{array}{cc} {1+1} & {5+(-1)} \\ {-1+5} & {2+2} \end{array}\right]$
$\Rightarrow A+A^{\prime}=\left[\begin{array}{ll} {2} & {4} \\ {4} & {4} \end{array}\right]$
Now, Let M = $\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$
Therefore, M = $\frac{1}{2}\left[\begin{array}{ll} {2} & {4} \\ {4} & {4} \end{array}\right]$
$\Rightarrow \mathrm{M}=\left[\begin{array}{ll} {1} & {2} \\ {2} & {2} \end{array}\right]$
Now, M' = $\left[\begin{array}{ll} {1} & {2} \\ {2} & {2} \end{array}\right]$
$\Rightarrow \mathrm{M}^{\prime}=\mathrm{M}$
Thus M = $\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$ is a symmetric matrix as M' = M
Now, on subtracting A’ from A we will get,
$A-A^{\prime}=\left[\begin{array}{cc} {1} & {5} \\ {-1} & {2} \end{array}\right]-\left[\begin{array}{cc} {1} & {-1} \\ {5} & {2} \end{array}\right]$
$\Rightarrow A-A^{\prime}=\left[\begin{array}{cc} {1-1} & {5-(-1)} \\ {-1-5} & {2-2} \end{array}\right]$
$\Rightarrow A-A^{\prime}=\left[\begin{array}{cc} {0} & {6} \\ {-6} & {0} \end{array}\right]$
Now, Let N = $\frac{1}{2}\left(\mathrm{A}-\mathrm{A}^{\prime}\right)$
Therefore, N = $\frac{1}{2}\left[\begin{array}{cc} {0} & {6} \\ {-6} & {0} \end{array}\right]$
$\Rightarrow N=\left[\begin{array}{cc} {0} & {3} \\ {-3} & {0} \end{array}\right]$
Now, N' = $\left[\begin{array}{cc} {0} & {-3} \\ {3} & {0} \end{array}\right]$
$\Rightarrow \mathrm{N}^{\prime}=(-1)\left[\begin{array}{cc} {0} & {3} \\ {-3} & {0} \end{array}\right]$
$\Rightarrow \mathrm{N}^{\prime}=-\mathrm{N}$
Thus M = $\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$ is a skew-symmetric matrix as N' = - N
Now, Add M and N, we get,
$M+N=\left[\begin{array}{ll} {1} & {2} \\ {2} & {2} \end{array}\right]+\left[\begin{array}{cc} {0} & {3} \\ {-3} & {0} \end{array}\right]$
$\Rightarrow \mathrm{M}+\mathrm{N}=\left[\begin{array}{cc} {1+0} & {2+3} \\ {2+(-3)} & {2+0} \end{array}\right]$
$\Rightarrow \mathrm{M}+\mathrm{N}=\left[\begin{array}{cc} {1} & {5} \\ {-1} & {2} \end{array}\right]$
So we see here, $M+N=\left[\begin{array}{cc} {1} & {5} \\ {-1} & {2} \end{array}\right]=A$
Thus, A is represented as the sum of a symmetric matrix M and a skew-symmetric matrix N.
View full question & answer→Question 152 Marks
Express matrix as the sum of a symmetric and a skew-symmetric matrix:$\left[\begin{array}{ccc} {3} & {3} & {-1} \\ {-2} & {-2} & {1} \\ {-4} & {-5} & {2} \end{array}\right]$
AnswerAs per Theorem “Any square matrix can be expressed as the sum of a symmetric and skew-symmetric matrix.” So in order to prove this, we will be using Theorem which states that “For any square matrix A with real number entries, A + A’ is a symmetric matrix and A – A’ is a skew-symmetric matrix.”
Now, Let A = $\left[\begin{array}{ccc} {3} & {3} & {-1} \\ {-2} & {-2} & {1} \\ {-4} & {-5} & {2} \end{array}\right]$ Therefore, A' = $\left[\begin{array}{ccc} {3} & {-2} & {-4} \\ {3} & {-2} & {-5} \\ {-1} & {1} & {2} \end{array}\right]$
Now, on adding A and A’ we will get,
$A+A^{\prime}=\left[\begin{array}{ccc} {3} & {3} & {-1} \\ {-2} & {-2} & {1} \\ {-4} & {-5} & {2} \end{array}\right]+\left[\begin{array}{ccc} {3} & {-2} & {-4} \\ {3} & {-2} & {-5} \\ {-1} & {1} & {2} \end{array}\right]$
$\Rightarrow A+A^{\prime}=\left[\begin{array}{ccc} {3+3} & {3+(-2)} & {-1+(-4)} \\ {-2+3} & {-2+(-2)} & {1+(-5)} \\ {-4+(-1)} & {-5+1} & {2+2} \end{array}\right]$
$\Rightarrow A+A^{\prime}=\left[\begin{array}{rrr} {6} & {1} & {-5} \\ {1} & {-4} & {-4} \\ {-5} & {-4} & {4} \end{array}\right]$
Now, Let M = $\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$
Therefore, M = $\frac{1}{2}\left[\begin{array}{ccc} {6} & {1} & {-5} \\ {1} & {-4} & {-4} \\ {-5} & {-4} & {4} \end{array}\right]$
$\Rightarrow \mathrm{M}=\left[\begin{array}{ccc} {3} & {\frac{1}{2}} & {\frac{-5}{2}} \\ {\frac{1}{2}} & {-2} & {-2} \\ {\frac{-5}{2}} & {-2} & {2} \end{array}\right]$
Now, $\mathrm{M}^{\prime}=\left[\begin{array}{ccc} {3} & {\frac{1}{2}} & {\frac{-5}{2}} \\ {\frac{1}{2}} & {-2} & {-2} \\ {\frac{-5}{2}} & {-2} & {2} \end{array}\right]$
$\Rightarrow$ M' = M
Thus M = $\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$ is a symmetric matrix as M' = M
Now, on subtracting A’ from A we will get,
$A-A^{\prime}=\left[\begin{array}{ccc} {3} & {3} & {-1} \\ {-2} & {-2} & {1} \\ {-4} & {-5} & {2} \end{array}\right]-\left[\begin{array}{ccc} {3} & {-2} & {-4} \\ {3} & {-2} & {-5} \\ {-1} & {1} & {2} \end{array}\right]$
$\Rightarrow A-A^{\prime}=\left[\begin{array}{ccc} {3-3} & {3-(-2)} & {-1-(-4)} \\ {-2-3} & {-2-(-2)} & {1-(-5)} \\ {-4-(-1)} & {-5-1} & {2-2} \end{array}\right]$
$\Rightarrow A-A^{\prime}=\left[\begin{array}{ccc} {0} & {5} & {3} \\ {-5} & {0} & {6} \\ {-3} & {-6} & {0} \end{array}\right]$
Now, Let N = $\frac{1}{2}\left(\mathrm{A}-\mathrm{A}^{\prime}\right)$
Therefore, N = $\frac{1}{2}\left[\begin{array}{ccc} {0} & {5} & {3} \\ {-5} & {0} & {6} \\ {-3} & {-6} & {0} \end{array}\right]$
$\Rightarrow \mathrm{N}=\left[\begin{array}{ccc} {0} & {\frac{5}{2}} & {\frac{3}{2}} \\ {\frac{-5}{2}} & {0} & {3} \\ {\frac{-3}{2}} & {-3} & {0} \end{array}\right]$
Now, $\mathrm{N}^{\prime}=\left[\begin{array}{ccc} {0} & {\frac{-5}{2}} & {\frac{-3}{2}} \\ {\frac{5}{2}} & {0} & {-3} \\ {\frac{3}{2}} & {3} & {0} \end{array}\right]$
$\Rightarrow \mathrm{N}^{\prime}=-\mathrm{N}$
Thus M = $\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$ is a symmetric matrix as N' = - N
w, Add M and N, we get,
$M+N=\left[\begin{array}{ccc} {3} & {\frac{1}{2}} & {\frac{-5}{2}} \\ {\frac{1}{2}} & {-2} & {-2} \\ {\frac{-5}{2}} & {-2} & {2} \end{array}\right]+\left[\begin{array}{ccc} {0} & {\frac{5}{2}} & {\frac{3}{2}} \\ {\frac{-5}{2}} & {0} & {3} \\ {\frac{-3}{2}} & {-3} & {0} \end{array}\right]$
$\Rightarrow \mathrm{M}+\mathrm{N}=\left[\begin{array}{ccc} {3+0} & {\frac{1}{2}+\frac{5}{2}} & {\frac{-5}{2}+\frac{3}{2}} \\ {\frac{1}{2}+\frac{-5}{2}} & {-2+0} & {-2+3} \\ {\frac{-5}{2}+\frac{-3}{2}} & {-2+(-3)} & {0+2} \end{array}\right]$
$\Rightarrow \mathrm{M}+\mathrm{N}=\left[\begin{array}{ccc} {3} & {\frac{6}{2}} & {\frac{-2}{2}} \\ {\frac{-4}{2}} & {-2} & {1} \\ {\frac{-8}{2}} & {-5} & {2} \end{array}\right]$
So we see here, M + N = $\left[\begin{array}{ccc} {3} & {3} & {-1} \\ {-2} & {-2} & {1} \\ {-4} & {-5} & {2} \end{array}\right]=A$
Thus, A is represented as the sum of a symmetric matrix M and a skew-symmetric matrix N.
View full question & answer→Question 162 Marks
Express matrix as the sum of a symmetric and a skew-symmetric matrix: $\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]$
AnswerAs per Theorem “Any square matrix can be expressed as the sum of a symmetric and skew-symmetric matrix.” So in order to prove this, we will be using Theorem which states that “For any square matrix A with real number entries, A + A’ is a symmetric matrix and A – A’ is a skew-symmetric matrix.”
Now, Let A = $\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]$
Therefore, A' = $\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]$
Now, on adding A and A’ we will get,
$A+A^{\prime}=\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]+\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]$
$\Rightarrow A+A^{\prime}=\left[\begin{array}{rrr} {6+6} & {-2+(-2)} & {2+2} \\ {-2+(-2)} & {3+3} & {-1+(-1)} \\ {2+2} & {-1+(-1)} & {3+3} \end{array}\right]$
$\Rightarrow \mathrm{A}+\mathrm{A}^{\prime}=\left[\begin{array}{ccc} {12} & {-4} & {4} \\ {-4} & {6} & {-2} \\ {4} & {-2} & {6} \end{array}\right]$
Let, M = $\frac{1}{2}\left[\begin{array}{ccc} {12} & {-4} & {4} \\ {-4} & {6} & {-2} \\ {4} & {-2} & {6} \end{array}\right]$
$\Rightarrow M=\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]$
Now, $M^{\prime}=\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]$ $\Rightarrow$ M' = M
Thus M = $\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$ is a symmetric matrix as M' = M
Now, on subtracting A’ from A we will get,
$A-A^{\prime}=\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]-\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]$
$\Rightarrow \mathrm{A}-\mathrm{A}^{\prime}=\left[\begin{array}{ccc} {6-6} & {-2-(-2)} & {2-2} \\ {-2} & {3-3} & {-1-(-1)} \\ {2-2} & {-1-(-1)} & {3-3} \end{array}\right]$
$\Rightarrow A-A^{\prime}=\left[\begin{array}{lll} {0} & {0} & {0} \\ {0} & {0} & {0} \\ {0} & {0} & {0} \end{array}\right]$
Now, Let N = $\frac{1}{2}\left(\mathrm{A}-\mathrm{A}^{\prime}\right)$
Therefore, N = $\frac{1}{2}\left[\begin{array}{lll} {0} & {0} & {0} \\ {0} & {0} & {0} \\ {0} & {0} & {0} \end{array}\right]$
$\Rightarrow \mathrm{N}=\left[\begin{array}{lll} {0} & {0} & {0} \\ {0} & {0} & {0} \\ {0} & {0} & {0} \end{array}\right]$
Now, $N^{\prime}=\left[\begin{array}{lll} {0} & {0} & {0} \\ {0} & {0} & {0} \\ {0} & {0} & {0} \end{array}\right]$
$\Rightarrow \mathrm{N}^{\prime}=(-1)\left[\begin{array}{lll} {0} & {0} & {0} \\ {0} & {0} & {0} \\ {0} & {0} & {0} \end{array}\right]$
$\Rightarrow$ N’ = - N
Thus M = $\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$ is a symmetric matrix as N' = - N.
Now, Add M and N, we get,
$M+N=\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]+\left[\begin{array}{lll} {0} & {0} & {0} \\ {0} & {0} & {0} \\ {0} & {0} & {0} \end{array}\right]$
$\Rightarrow \mathrm{M}+\mathrm{N}=\left[\begin{array}{ccc} {6+0} & {-2+0} & {2+0} \\ {-2+0} & {3+0} & {-1+0} \\ {2+0} & {-1+0} & {3+0} \end{array}\right]$
$\Rightarrow \mathrm{M}+\mathrm{N}=\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]$ So we see her, M + N = $\left[\begin{array}{ccc} {6} & {-2} & {2} \\ {-2} & {3} & {-1} \\ {2} & {-1} & {3} \end{array}\right]$ = A
Thus, A is represented as the sum of a symmetric matrix M and a skew-symmetric matrix N.
View full question & answer→Question 172 Marks
Express matrix as the sum of a symmetric and a skew-symmetric matrix:$\left[\begin{array}{rr} {3} & {5} \\ {1} & {-1} \end{array}\right]$
AnswerAs per Theorem “Any square matrix can be expressed as the sum of a symmetric and skew-symmetric matrix.” So in order to prove this, we will be using Theorem which states that “For any square matrix A with real number entries, A + A’ is a symmetric matrix and A – A’ is a skew-symmetric matrix.”
Now, Let A = $\left[\begin{array}{cc} {3} & {5} \\ {1} & {-1} \end{array}\right]$
Therefore, A' = $\left[\begin{array}{cc} {3} & {1} \\ {5} & {-1} \end{array}\right]$
Now, on adding A and A’ we will get,
$A+A^{\prime}=\left[\begin{array}{cc} {3} & {5} \\ {1} & {-1} \end{array}\right]+\left[\begin{array}{cc} {3} & {1} \\ {5} & {-1} \end{array}\right]$
$\Rightarrow \mathrm{A}+\mathrm{A}^{\prime}=\left[\begin{array}{cc} {3+3} & {5+1} \\ {1+5} & {-1+(-1)} \end{array}\right]$
$\Rightarrow A+A^{\prime}=\left[\begin{array}{cc} {6} & {6} \\ {6} & {-2} \end{array}\right]$
Now, Let M = $\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$
Therefore, $M=\frac{1}{2}\left[\begin{array}{cc} {6} & {6} \\ {6} & {-2} \end{array}\right] = \left[\begin{array}{cc} {3} & {3} \\ {3} & {-1} \end{array}\right]$
Now, $M^{\prime}=\left[\begin{array}{cc} {3} & {3} \\ {3} & {-1} \end{array}\right]$
$\Rightarrow$ M' = M Thus M = $\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$ is a symmetric matrix as M' = M.
Now on subtracting A’ from A we will get,
$A-A^{\prime}=\left[\begin{array}{cc} {3} & {5} \\ {1} & {-1} \end{array}\right]-\left[\begin{array}{cc} {3} & {1} \\ {5} & {-1} \end{array}\right]$
$\Rightarrow A-A^{\prime}=\left[\begin{array}{cc} {0} & {4} \\ {-4} & {0} \end{array}\right]$
Now, Let N = $\frac{1}{2}\left(\mathrm{A}-\mathrm{A}^{\prime}\right)$
Therefore, $N=\frac{1}{2}\left[\begin{array}{cc} {0} & {4} \\ {-4} & {0} \end{array}\right]$
$\Rightarrow \mathrm{N}=\left[\begin{array}{cc} {0} & {2} \\ {-2} & {0} \end{array}\right]$
Now, N' =$\left[\begin{array}{cc} {0} & {-2} \\ {2} & {0} \end{array}\right]$
$\Rightarrow \mathrm{N}^{\prime}=(-1)\left[\begin{array}{cc} {0} & {2} \\ {-2} & {0} \end{array}\right]$
$\Rightarrow$ N' = - N
Thus M = $\frac{1}{2}\left(\mathrm{A}+\mathrm{A}^{\prime}\right)$ is a skew-symmetric matrix as N' = -N
Now, Add M and N, we get,
$M+N=\left[\begin{array}{cc} {3} & {3} \\ {3} & {-1} \end{array}\right]+\left[\begin{array}{cc} {0} & {2} \\ {-2} & {0} \end{array}\right]$
$\Rightarrow \mathrm{M}+\mathrm{N}=\left[\begin{array}{cc} {3+0} & {3+2} \\ {3+(-2)} & {-1+0} \end{array}\right]$
$\Rightarrow \mathrm{M}+\mathrm{N}=\left[\begin{array}{cc} {3} & {5} \\ {1} & {-1} \end{array}\right]$
So we see here, M + N = $\left[\begin{array}{cc} {3} & {5} \\ {1} & {-1} \end{array}\right]$ = A
Thus, A is represented as the sum of a symmetric matrix M and a skew-symmetric matrix N.
Hence proved.
View full question & answer→Question 182 Marks
Find $x$ and $y$, if $2\left[\begin{array}{ll}1 & 3 \\ 0 & x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right]$.
AnswerGiven: $2\left[ {\begin{array}{*{20}{c}} 1&3 \\ 0&x \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} y&0 \\ 1&2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5&6 \\ 1&8 \end{array}} \right]$
$\Rightarrow \left[ {\begin{array}{*{20}{c}} 2&6 \\ 0&{2x} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} y&0 \\ 1&2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5&6 \\ 1&8 \end{array}} \right]$
$\Rightarrow \left[ {\begin{array}{*{20}{c}} {2 + y}&6 \\ 1&{2x + x} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5&6 \\ 1&8 \end{array}} \right]$
Equating corresponding entries, we have
2 + y = 5 and 2x + 2 = 8
$\Rightarrow$ y = 5 - 2 and 2(x + 1) = 8
$\Rightarrow$ y = 3 and x + 1 = 4
$\Rightarrow$ y = 3 and x = 3
View full question & answer→Question 192 Marks
Find X, if Y = $\left[\begin{array}{ll} {3} & {2} \\ {1} & {4} \end{array}\right] \text { and } 2 \mathrm{X}+\mathrm{Y}=\left[\begin{array}{rr} {1} & {0} \\ {-3} & {2} \end{array}\right]$
Answer2X + Y = $\left[\begin{array}{cc} {1} & {0} \\ {-3} & {2} \end{array}\right]$
$\Rightarrow 2 \mathrm{X}+\left[\begin{array}{ll} {3} & {2} \\ {1} & {4} \end{array}\right]=\left[\begin{array}{cc} {1} & {0} \\ {-3} & {2} \end{array}\right]$
$\Rightarrow 2 \mathrm{X}=\left[\begin{array}{cc} {1} & {0} \\ {-3} & {2} \end{array}\right]-\left[\begin{array}{cc} {3} & {2} \\ {1} & {4} \end{array}\right]$
$\Rightarrow 2 \mathrm{X}=\left[\begin{array}{cc} {1-3} & {0-2} \\ {-3-1} & {2-4} \end{array}\right]$
$\Rightarrow 2 \mathrm{X}=\left[\begin{array}{cc} {-2} & {-2} \\ {-4} & {-2} \end{array}\right]$
$\Rightarrow X=\frac{1}{2}\left[\begin{array}{ll} {-2} & {-2} \\ {-4} & {-2} \end{array}\right]$
$\Rightarrow X=\left[\begin{array}{cc} {-1} & {-1} \\ {-2} & {-1} \end{array}\right]$
View full question & answer→Question 202 Marks
Find X and Y, if $2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{ll} {2} & {3} \\ {4} & {0} \end{array}\right] \text { and } 3 \mathrm{X}+2 \mathrm{Y}=\left[\begin{array}{cc} {2} & {-2} \\ {-1} & {5} \end{array}\right]$
Answer2X + 3Y = $\left[\begin{array}{ll} {2} & {3} \\ {4} & {0} \end{array}\right]$ ......(1)
3X + 2Y = $\left[\begin{array}{cc} {2} & {-2} \\ {-1} & {5} \end{array}\right]$ .....(2)
Now, multiply equation (1) by 2 and equation (2) by 3, we get,
4X + 6Y = $\left[\begin{array}{ll} {4} & {6} \\ {8} & {0} \end{array}\right]$ .....(3)
9X + 6Y = $\left[\begin{array}{cc} {6} & {-6} \\ {-3} & {15} \end{array}\right]$ ....(4)
Subtracting equation (4) from (3), we get,
(4X + 6Y) – (9X + 6Y) = $\left[\begin{array}{cc} {4} & {6} \\ {8} & {0} \end{array}\right]-\left[\begin{array}{cc} {6} & {-6} \\ {-3} & {15} \end{array}\right]$
$\Rightarrow-5 \mathrm{X}=\left[\begin{array}{cc} {4-6} & {6-(-6)} \\ {8-(-3)} & {0-15} \end{array}\right]$
$=\left[\begin{array}{cc} {-2} & {12} \\ {11} & {-15} \end{array}\right]$
$\Rightarrow \mathrm{X}=-\frac{1}{5}\left[\begin{array}{cc} {-2} & {12} \\ {11} & {-15} \end{array}\right]=\left[\begin{array}{cc} {\frac{2}{5}} & {\frac{-12}{5}} \\ {\frac{-11}{5}} & {3} \end{array}\right]$
Now, 2X + 3Y = $\left[\begin{array}{ll} {2} & {3} \\ {4} & {0} \end{array}\right]$
$\Rightarrow 2\left[\begin{array}{cc} {\frac{2}{5}} & {\frac{-12}{5}} \\ {\frac{-11}{5}} & {3} \end{array}\right]+3 \mathrm{Y}=\left[\begin{array}{cc} {2} & {3} \\ {4} & {0} \end{array}\right]$
$\Rightarrow\left[\begin{array}{cc} {\frac{4}{5}} & {\frac{-24}{5}} \\ {\frac{-22}{5}} & {6} \end{array}\right]+3 \mathrm{Y}=\left[\begin{array}{ll} {2} & {3} \\ {4} & {0} \end{array}\right]$
$\Rightarrow 3 \mathrm{Y}=\left[\begin{array}{ll} {2} & {3} \\ {4} & {0} \end{array}\right]-\left[\begin{array}{cc} {\frac{4}{5}} & {\frac{-24}{5}} \\ {\frac{-22}{5}} & {6} \end{array}\right]$
$\Rightarrow 3 \mathrm{Y}=\left[\begin{array}{cc} {2-\frac{4}{5}} & {3+\frac{24}{5}} \\ {4+\frac{22}{5}} & {0-6} \end{array}\right]=\left[\begin{array}{cc} {\frac{6}{5}} & {\frac{39}{5}} \\ {\frac{42}{5}} & {-6} \end{array}\right]$
$\Rightarrow Y=\frac{1}{3}\left[\begin{array}{cc} {\frac{6}{5}} & {\frac{39}{5}} \\ {\frac{42}{5}} & {-6} \end{array}\right]$
$\Rightarrow Y=\left[\begin{array}{cc} {\frac{2}{5}} & {\frac{13}{5}} \\ {\frac{14}{5}} & {-2} \end{array}\right]$
View full question & answer→Question 212 Marks
Find X and Y, if $\mathrm{X}+\mathrm{Y}=\left[\begin{array}{ll} {7} & {0} \\ {2} & {5} \end{array}\right] \text { and } \mathrm{X}-\mathrm{Y}=\left[\begin{array}{ll} {3} & {0} \\ {0} & {3} \end{array}\right]$
AnswerGiven that, X + Y = $\left[\begin{array}{ll} {7} & {0} \\ {2} & {5} \end{array}\right]$ .....(1)
X - Y = $\left[\begin{array}{ll} {3} & {0} \\ {0} & {3} \end{array}\right]$ ....(2)
Adding (1) and (2), we get
$2 X=\left[\begin{array}{ll} {7} & {0} \\ {2} & {5} \end{array}\right]+\left[\begin{array}{ll} {3} & {0} \\ {0} & {3} \end{array}\right]$
$=\left[\begin{array}{ll} {7+3} & {0+0} \\ {2+0} & {5+3} \end{array}\right]$
$=\left[\begin{array}{cc} {10} & {0} \\ {2} & {8} \end{array}\right]$
$\Rightarrow X=\frac{1}{2}\left[\begin{array}{ll} {10} & {0} \\ {2} & {8} \end{array}\right]=\left[\begin{array}{ll} {5} & {0} \\ {1} & {4} \end{array}\right]$
Now, X + Y = $\left[\begin{array}{ll} {7} & {0} \\ {2} & {5} \end{array}\right]$
$\Rightarrow\left[\begin{array}{ll} {5} & {0} \\ {1} & {4} \end{array}\right]+Y=\left[\begin{array}{ll} {7} & {0} \\ {2} & {5} \end{array}\right]$
$\Rightarrow Y=\left[\begin{array}{ll} {7} & {0} \\ {2} & {5} \end{array}\right]-\left[\begin{array}{ll} {5} & {0} \\ {1} & {4} \end{array}\right]$
$\Rightarrow Y=\left[\begin{array}{ll} {2} & {0} \\ {1} & {1} \end{array}\right]$
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Simplify $\cos \theta \left[ \begin{array} { c c } { \cos \theta } & { \sin \theta } \\ { - \sin \theta } & { \cos \theta } \end{array} \right] + \sin \theta \left[ \begin{array} { c c } { \sin \theta } & { - \cos \theta } \\ { \cos \theta } & { \sin \theta } \end{array} \right]$.
AnswerAccording to the question, $\cos \theta \left[ \begin{array} { c c } { \cos \theta } & { \sin \theta } \\ { - \sin \theta } & { \cos \theta } \end{array} \right] + \sin \theta \left[ \begin{array} { c c } { \sin \theta } & { - \cos \theta } \\ { \cos \theta } & { \sin \theta } \end{array} \right]$
$= \left[ \begin{array} { c c } { \cos ^ { 2 } \theta } & { \sin \theta \cos \theta } \\ { - \sin \theta \cos \theta } & { \cos ^ { 2 } \theta } \end{array} \right] + \left[ \begin{array} { c c } { \sin ^ { 2 } \theta } & { - \sin \theta \cos \theta } \\ { \sin \theta \cos \theta } & { \sin ^ { 2 } \theta } \end{array} \right]$
$= \left[ \begin{array} { c c } { \cos ^ { 2 } \theta + \sin ^ { 2 } \theta } & { \sin \theta \cos \theta - \sin \theta \cos \theta } \\ { - \sin \theta \cos \theta + \sin \theta \cos \theta } & { \cos ^ { 2 } \theta + \sin ^ { 2 } \theta } \end{array} \right]$
$ = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&{ 1} \end{array}} \right]\quad \left[ {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right]$
= I = unit matrix.
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If $A=\left[\begin{array}{ccc}\frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3}\end{array}\right]$ and $B=\left[\begin{array}{ccc}\frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5}\end{array}\right]$, then compute $3 A-5 B$.
Answer3A – 5B = $3\left[ {\begin{array}{*{20}{c}} {\frac{2}{3}}&1&{\frac{5}{3}} \\ {\frac{1}{3}}&{\frac{2}{3}}&{\frac{4}{3}} \\ {\frac{7}{3}}&2&{\frac{2}{3}} \end{array}} \right] - 5\left[ {\begin{array}{*{20}{c}} {\frac{2}{5}}&{\frac{3}{5}}&1 \\ {\frac{1}{5}}&{\frac{2}{5}}&{\frac{4}{5}} \\ {\frac{7}{5}}&{\frac{6}{5}}&{\frac{2}{5}} \end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}} 2&3&5 \\ 1&2&4 \\ 7&6&2 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 2&3&5 \\ 1&2&4 \\ 7&6&2 \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {2 - 2}&{3 - 3}&{5 - 5} \\ {1 - 1}&{2 - 2}&{4 - 4} \\ {7 - 7}&{6 - 6}&{2 - 2} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{array}} \right]$
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Compute the indicated products.$\left[\begin{array}{rrr} {3} & {-1} & {3} \\ {-1} & {0} & {2} \end{array}\right]\left[\begin{array}{rr} {2} & {-3} \\ {1} & {0} \\ {3} & {1} \end{array}\right]$
Answer$\left[\begin{array}{ccc} {3} & {-1} & {3} \\ {-1} & {0} & {2} \end{array}\right]\left[\begin{array}{cc} {2} & {-3} \\ {1} & {0} \\ {3} & {1} \end{array}\right]$
= $\left[\begin{array}{cc} {3(2)-1(1)+3(3)} & {3(-3)-1(0)+3(1)} \\ {-1(2)+0(1)+2(3)} & {-1(-3)+0(0)+2(1)} \end{array}\right]$
= $\left[\begin{array}{cc} {6-1+9} & {-9-0+3} \\ {-2+0+6} & {3+0+2} \end{array}\right]$
= $\left[\begin{array}{cc} {14} & {-6} \\ {4} & {5} \end{array}\right]$
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Compute the indicated products. $\left[\begin{array}{cc} {2} & {1} \\ {3} & {2} \\ {-1} & {1} \end{array}\right]\left[\begin{array}{ccc} {1} & {0} & {1} \\ {-1} & {2} & {1} \end{array}\right]$
AnswerWe have, $\left[\begin{array}{cc} {2} & {1} \\ {3} & {2} \\ {-1} & {1} \end{array}\right]\left[\begin{array}{ccc} {1} & {0} & {1} \\ {-1} & {2} & {1} \end{array}\right]$
= $\left[\begin{array}{ccc} {2(1)+1(-1)} & {2(0)+1(2)} & {2(1)+1(1)} \\ {3(1)+2(-1)} & {3(0)+2(2)} & {3(1)+2(1)} \\ {-1(1)+1(-1)} & {-1(0)+1(2)} & {-1(1)+1(1)} \end{array}\right]$
= $\left[\begin{array}{ccc} {2-1} & {0+2} & {2+1} \\ {3-2} & {0+4} & {3+2} \\ {-1-1} & {0+2} & {-1+1} \end{array}\right]$
= $\left[\begin{array}{ccc} {1} & {2} & {3} \\ {1} & {4} & {5} \\ {-2} & {2} & {0} \end{array}\right]$
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Compute the indicated product $\left[\begin{array}{ccc} {2} & {3} & {4} \\ {3} & {4} & {5} \\ {4} & {5} & {6} \end{array}\right]\left[\begin{array}{rrr} {1} & {-3} & {5} \\ {0} & {2} & {4} \\ {3} & {0} & {5} \end{array}\right]$
AnswerWe have, $\left[\begin{array}{lll} {2} & {3} & {4} \\ {3} & {4} & {5} \\ {4} & {5} & {6} \end{array}\right]\left[\begin{array}{rrr} {1} & {-3} & {5} \\ {0} & {2} & {4} \\ {3} & {0} & {5} \end{array}\right]$
= $\left[\begin{array}{ccc} {2(1)+3(0)+4(3)} & {2(-3)+3(2)+4(0)} & {2(5)+3(4)+4(5)} \\ {3(1)+4(0)+5(3)} & {3(-3)+4(2)+5(0)} & {3(5)+4(4)+5(5)} \\ {4(1)+5(0)+6(3)} & {4(-3)+5(2)+6(0)} & {4(5)+5(4)+6(5)} \end{array}\right]$
= $\left[\begin{array}{ccc} {2+0+12} & {-6+6+0} & {10+12+20} \\ {3+0+15} & {-9+8+0} & {15+16+25} \\ {4+0+18} & {-12+10+0} & {20+20+30} \end{array}\right]$
= $\left[\begin{array}{ccc} {14} & {0} & {42} \\ {18} & {-1} & {56} \\ {22} & {-2} & {70} \end{array}\right]$
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Compute the indicated product $\left[\begin{array}{rr} {1} & {-2} \\ {2} & {3} \end{array}\right]\left[\begin{array}{lll} {1} & {2} & {3} \\ {2} & {3} & {1} \end{array}\right]$
Answer$\left[\begin{array}{cc} {1} & {-2} \\ {2} & {3} \end{array}\right]\left[\begin{array}{lll} {1} & {2} & {3} \\ {3} & {2} & {1} \end{array}\right]$
= $\left[\begin{array}{lll} {1(1)-2(2)} & {1(2)-2(3)} & {1(3)-2(1)} \\ {2(1)+3(2)} & {2(2)+3(3)} & {2(3)+3(1)} \end{array}\right]$
= $\left[\begin{array}{ccc} {1-4} & {2-6} & {3-2} \\ {2+6} & {4+9} & {6+3} \end{array}\right]$
= $\left[\begin{array}{ccc} {-3} & {-4} & {1} \\ {8} & {13} & {9} \end{array}\right]$
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Compute the indicated product $\left[\begin{array}{l} {1} \\ {2} \\ {3} \end{array}\right]\left[\begin{array}{lll} {2} & {3} & {4} \end{array}\right]$
Answer$\left[\begin{array}{l} {1} \\ {2} \\ {3} \end{array}\right]\left[\begin{array}{lll} {2} & {3} & {4} \end{array}\right]$
$= \left[\begin{array}{ccc} {1(2)} & {1(3)} & {1(4)} \\ {2(2)} & {2(3)} & {2(4)} \\ {3(2)} & {3(3)} & {3(4)} \end{array}\right]$
$= \left[\begin{array}{ccc} {2} & {3} & {4} \\ {4} & {6} & {8} \\ {6} & {9} & {12} \end{array}\right]$
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Compute the indicated product
$\left[\begin{array}{cc} {a} & {b} \\ {-b} & {a} \end{array}\right]\left[\begin{array}{cc} {a} & {-b} \\ {b} & {a} \end{array}\right]$
AnswerAccording to the Question,
$\left[\begin{array}{cc} {a} & {b} \\ {-b} & {a} \end{array}\right]\left[\begin{array}{cc} {a} & {-b} \\ {b} & {a} \end{array}\right]$
= $\left[\begin{array}{cc} {a(a)+b(b)} & {a(-b)+b(a)} \\ {-b(a)+a(b)} & {-b(-b)+a(a)} \end{array}\right]$
= $\left[\begin{array}{cc} {a^{2}+b^{2}} & {-a b+a b} \\ {-a b+a b} & {b^{2}+a^{2}} \end{array}\right]$
= $\left[\begin{array}{cc} {a^{2}+b^{2}} & {0} \\ {0} & {b^{2}+a^{2}} \end{array}\right]$
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Compute $\left[\begin{array}{cc} {\cos ^{2} x} & {\sin ^{2} x} \\ {\sin ^{2} x} & {\cos ^{2} x} \end{array}\right]+\left[\begin{array}{cc} {\sin ^{2} x} & {\cos ^{2} x} \\ {\cos ^{2} x} & {\sin ^{2} x} \end{array}\right]$
Answer$\left[\begin{array}{cc} {\cos ^{2} x} & {\sin ^{2} x} \\ {\sin ^{2} x} & {\cos ^{2} x} \end{array}\right]+\left[\begin{array}{cc} {\sin ^{2} x} & {\cos ^{2} x} \\ {\cos ^{2} x} & {\sin ^{2} x} \end{array}\right]$
=$\left[\begin{array}{cc} {\cos ^{2} x+\sin ^{2} x} & {\cos ^{2} x+\sin ^{2} x} \\ {\sin ^{2} x+\cos ^{2} x} & {\cos ^{2} x+\sin ^{2} x} \end{array}\right]$
= $\left[\begin{array}{ll} {1} & {1} \\ {1} & {1} \end{array}\right]$
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Compute $\left[\begin{array}{ll} {a^{2}+b^{2}} & {b^{2}+c^{2}} \\ {a^{2}+c^{2}} & {a^{2}+b^{2}} \end{array}\right]+\left[\begin{array}{cc} {2 a b} & {2 b c} \\ {-2 a c} & {-2 a b} \end{array}\right]$
Answer$\left[\begin{array}{ll} {a^{2}+b^{2}} & {b^{2}+c^{2}} \\ {a^{2}+c^{2}} & {a^{2}+b^{2}} \end{array}\right]+\left[\begin{array}{cc} {2 a b} & {2 b c} \\ {-2 a c} & {-2 a b} \end{array}\right]$
$=\left[\begin{array}{ll} {a^{2}+b^{2}+2 a b} & {b^{2}+c^{2}+2 b c} \\ {a^{2}+c^{2}-2 a c} & {a^{2}+b^{2}-2 a b} \end{array}\right]$
$=\left[\begin{array}{ll} {(a+b)^{2}} & {(b+c)^{2}} \\ {(a-c)^{2}} & {(a-b)^{2}} \end{array}\right]$
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Compute $\left[\begin{array}{cc} {a} & {b} \\ {-b} & {a} \end{array}\right]+\left[\begin{array}{ll} {a} & {b} \\ {b} & {a} \end{array}\right]$
Answer$\left[\begin{array}{cc} {a} & {b} \\ {-b} & {a} \end{array}\right]+\left[\begin{array}{ll} {a} & {b} \\ {b} & {a} \end{array}\right]$
$=\left[\begin{array}{cc} {a+a} & {b+b} \\ {-b+b} & {a+a} \end{array}\right]$
$=\left[\begin{array}{cc} {2 a} & {2 b} \\ {0} & {2 a} \end{array}\right]$
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Show that $\left[\begin{array}{ccc} {1} & {2} & {3} \\ {0} & {1} & {0} \\ {1} & {1} & {0} \end{array}\right]\left[\begin{array}{rrr} {-1} & {1} & {0} \\ {0} & {-1} & {1} \\ {2} & {3} & {4} \end{array}\right] \neq\left[\begin{array}{rrr} {-1} & {1} & {0} \\ {0} & {-1} & {1} \\ {2} & {3} & {4} \end{array}\right]\left[\begin{array}{lll} {1} & {2} & {3} \\ {0} & {1} & {0} \\ {1} & {1} & {0} \end{array}\right]$
AnswerHere we have, $\left[\begin{array}{ccc} {1} & {2} & {3} \\ {0} & {1} & {0} \\ {1} & {1} & {0} \end{array}\right]\left[\begin{array}{ccc} {-1} & {1} & {0} \\ {0} & {-1} & {1} \\ {2} & {3} & {4} \end{array}\right]$
= $\left[\begin{array}{lll} {1(-1)+2(0)+3(2)} & {1(1)+2(-1)+3(3)} & {1(0)+2(1)+3(4)} \\ {0(-1)+1(0)+0(2)} & {0(1)+1(-1)+0(3)} & {0(0)+1(1)+0(4)} \\ {1(-1)+1(0)+0(2)} & {1(1)+1(-1)+0(3)} & {1(0)+1(1)+0(4)} \end{array}\right]$
= $\left[\begin{array}{ccc} {5} & {8} & {14} \\ {0} & {-1} & {1} \\ {-1} & {0} & {1} \end{array}\right]$
Now, $\left[\begin{array}{ccc} {-1} & {1} & {0} \\ {0} & {-1} & {1} \\ {2} & {3} & {4} \end{array}\right]\left[\begin{array}{lll} {1} & {2} & {3} \\ {0} & {1} & {0} \\ {1} & {1} & {0} \end{array}\right]$
= $\left[\begin{array}{ccc} {-1(1)+1(0)+0(1)} & {-1(2)+1(1)+0(1)} & {-1(3)+1(0)+0(0)} \\ {0(1)+(-1)(0)+1(1)} & {0(2)+(-1)(1)+1(1)} & {0(3)+)(-1)(0)+1(0)} \\ {2(1)+3(0)+4(1)} & {2(2)+3(1)+4(1)} & {2(3)+3(0)+4(0)} \end{array}\right]$
= $\left[\begin{array}{ccc} {-1} & {-1} & {-3} \\ {1} & {0} & {0} \\ {6} & {11} & {6} \end{array}\right]$
Therefore, $\left[\begin{array}{ccc} {1} & {2} & {3} \\ {0} & {1} & {0} \\ {1} & {1} & {0} \end{array}\right]\left[\begin{array}{ccc} {-1} & {1} & {0} \\ {0} & {-1} & {1} \\ {2} & {3} & {4} \end{array}\right] \neq\left[\begin{array}{ccc} {-1} & {1} & {0} \\ {0} & {-1} & {1} \\ {2} & {3} & {4} \end{array}\right]\left[\begin{array}{ccc} {1} & {2} & {3} \\ {0} & {1} & {0} \\ {1} & {1} & {0} \end{array}\right]$
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Show that $\left[\begin{array}{rr} {5} & {-1} \\ {6} & {7} \end{array}\right]\left[\begin{array}{ll} {2} & {1} \\ {3} & {4} \end{array}\right] \neq\left[\begin{array}{ll} {2} & {1} \\ {3} & {4} \end{array}\right]\left[\begin{array}{rr} {5} & {-1} \\ {6} & {7} \end{array}\right]$
AnswerL.H.S =$\left[\begin{array}{cc} {5} & {-1} \\ {6} & {7} \end{array}\right]\left[\begin{array}{ll} {2} & {1} \\ {3} & {4} \end{array}\right]$
= $\left[\begin{array}{cc} {5(2)-1(3)} & {5(1)-1(4)} \\ {6(2)+7(3)} & {6(1)+7(4)} \end{array}\right]$
= $\left[\begin{array}{cc} {10-3} & {5-4} \\ {12+21} & {6+28} \end{array}\right]$
= $\left[\begin{array}{cc} {7} & {1} \\ {33} & {34} \end{array}\right]~...(i)$
R.H.S = $\left[\begin{array}{ll} {2} & {1} \\ {3} & {4} \end{array}\right]\left[\begin{array}{ll} {5} & {-1} \\ {6} & {7} \end{array}\right]$
= $\left[\begin{array}{cc} {2(5)+1(6)} & {2(-1)+1(7)} \\ {3(5)+4(6)} & {3(-1)+4(7)} \end{array}\right]$
= $\left[\begin{array}{cc} {10+6} & {-2+7} \\ {15+24} & {-3+28} \end{array}\right]$
= $\left[\begin{array}{ll} {16} & {5} \\ {39} & {25} \end{array}\right]~...(ii)$
Therefore, from (i) and (ii), we get
$\left[\begin{array}{cc} {5} & {-1} \\ {6} & {7} \end{array}\right]\left[\begin{array}{ll} {2} & {1} \\ {3} & {4} \end{array}\right] \neq\left[\begin{array}{ll} {2} & {1} \\ {3} & {4} \end{array}\right]\left[\begin{array}{cc} {5} & {-1} \\ {6} & {7} \end{array}\right]$
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If $F(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$, show that $\mathrm{F}(\mathrm{x}) \mathrm{F}(\mathrm{y})=\mathrm{F}(\mathrm{x}+\mathrm{y})$
Answer$F(x) = \left[ {\begin{array}{*{20}{c}} {\cos x}&{ - \sin x}&0 \\ {\sin x}&{\cos x}&0 \\ 0&0&1 \end{array}} \right]$, $F(y) = \left[ {\begin{array}{*{20}{c}} {\cos y}&{ - \sin y}&0 \\ {\sin y}&{\cos y}&0 \\ 0&0&1 \end{array}} \right]$
$F(x + y) = \left[ {\begin{array}{*{20}{c}} {\cos (x + y)}&{ - \sin (x + y)}&0 \\ {\sin (x + y)}&{\cos (x + y)}&0 \\ 0&0&1 \end{array}} \right]$
F(x)F(y)$ = \left[ {\begin{array}{*{20}{c}} {\cos x}&{ - \sin x}&0 \\ {\sin x}&{\cos x}&0 \\ 0&0&1 \end{array}} \right]$$\left[ {\begin{array}{*{20}{c}} {\cos y}&{ - \sin y}&0 \\ {\sin y}&{\cos y}&0 \\ 0&0&1 \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {\cos x\cos y - \sin x\sin y + 0}&{ - \cos x\sin y - \sin x\cos y + 0}&0 \\ {\sin x\cos y + \cos x\sin y + 0}&{ - \sin x\sin y + \cos x\cos y + 0}&0 \\ 0&0&0 \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {\cos (x + y)}&{ - \sin (x + y)}&0 \\ {\sin (x + y)}&{\cos (x + y)}&0 \\ 0&0&1 \end{array}} \right]$
= F(x + y)
$\therefore$ F(x)F(y) = F(x + y)
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Given: $3\left[\begin{array}{cc}x & y \\ z & w\end{array}\right]=\left[\begin{array}{cc}x & 6 \\ -1 & 2 w\end{array}\right]+\left[\begin{array}{cc}4 & x+y \\ z+w & 3\end{array}\right]$, find the values of $\mathrm{x}, \mathrm{y}, \mathrm{z}$ and $\mathrm{w}$.
AnswerGiven: $3\left[ {\begin{array}{*{20}{c}} x&y \\ z&w \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} x&6 \\ { - 1}&{2w} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 4&{x + y} \\ {z + w}&3 \end{array}} \right]$
$\Rightarrow \left[ {\begin{array}{*{20}{c}} {3x}&{3y} \\ {3z}&{3w} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {x + 4}&{6 + x + y} \\ { - 1 + z + w}&{2w + 3} \end{array}} \right]$
Equating corresponding entries, we have
3x = x + 4 $\Rightarrow$ 2x = 4 $\Rightarrow$ x = 2
and 3y = 6 + x + y
$\Rightarrow$ 2y = 6 + 2
$\Rightarrow$ 2y = 8
$\Rightarrow$ y = 4
and 3z = -1 + z + w $\Rightarrow$ 2z - w = - 1 ….(i)
and 3w = 2w + 3 $\Rightarrow$ w = 3
Putting w = 3 in eq. (i), 2z - 3 = -1
$\Rightarrow$ 2z = 2 $\Rightarrow$ z = 1
$\therefore$ x = 2, y = 4, z = 1, w = 3
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If $x\left[\begin{array}{l}2 \\ 3\end{array}\right]+y\left[\begin{array}{c}-1 \\ 1\end{array}\right]=\left[\begin{array}{c}10 \\ 5\end{array}\right],$ find the values of $\mathrm{x}$ and $\mathrm{y}$.
AnswerGiven: $x\left[ {\begin{array}{*{20}{c}} 2 \\ 3 \end{array}} \right] + y\left[ {\begin{array}{*{20}{c}} { - 1} \\ 1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {10} \\ 5 \end{array}} \right]$
$\Rightarrow \left[ {\begin{array}{*{20}{c}} {2x} \\ {3x} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} { - y} \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {10} \\ 5 \end{array}} \right]$
$\Rightarrow \left[ {\begin{array}{*{20}{c}} {2x - y} \\ {3x + y} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {10} \\ 5 \end{array}} \right]$
Equating corresponding entries, we have
$2x - y = 10 ……….(i) $ and
$3x + y = 5 ……….(ii)$
Adding eq. $(i)$ and $(ii),$
we have $5x = 15$
$\Rightarrow x = 3$
Putting $x = 3$ in eq. $(ii),$
$9 + y = 5$
$\Rightarrow y = -4$
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Solve the equation for $\mathrm{x}, \mathrm{y}, \mathrm{z}$ and $\mathrm{t}$ if $2\left[\begin{array}{ll}x & z \\ y & t\end{array}\right]+3\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]=3\left[\begin{array}{cc}3 & 5 \\ 4 & 6\end{array}\right]$.
AnswerGiven: $2\left[ {\begin{array}{*{20}{c}} x&z \\ y&t \end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\ 0&2 \end{array}} \right] = 3\left[ {\begin{array}{*{20}{c}} 3&5 \\ 4&6 \end{array}} \right]$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}} {2x}&{2z} \\ {2y}&{2t} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 3&{ - 3} \\ 0&6 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 9&{15} \\ {12}&{18} \end{array}} \right]$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}} {2x + 3}&{2z - 3} \\ {2y + 0}&{2t + 6} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 9&{15} \\ {12}&{18} \end{array}} \right]$
Equating corresponding entries, we have
2x + 3 = 9 $\Rightarrow$ 2x = 9 - 3 $\Rightarrow$ 2x = 6 $\Rightarrow$ x = 3
and 2z - 3 = 15 $\Rightarrow$ 2z = 15 + 3 $\Rightarrow$ 2z = 18 $\Rightarrow$ z = 9
and 2y = 12 $\Rightarrow$ y = 6
And 2t + 6 = 18 $\Rightarrow$ 2t = 18 - 6 $\Rightarrow$ 2t = 12 $\Rightarrow$ t = 6
$\therefore$ x = 3, y = 6, z = 9, t = 6
View full question & answer→Question 392 Marks
Find the values of $\mathrm{a}, \mathrm{b}, \mathrm{c}$ and $\mathrm{d}$ from the equation $\left[\begin{array}{cc}a-b & 2 a+c \\ 2 a-b & 3 c+d\end{array}\right]=\left[\begin{array}{cc}-1 & 5 \\ 0 & 13\end{array}\right]$.
AnswerEquating corresponding entries,
a - b = -1 ……….(i)
2a - b = 0 ……….(ii)
2a + c = 5 ……….(iii)
3c + d = 13 ……….(iv)
Eq. (i) – Eq. (ii) = -a = -1
$\Rightarrow$ a = 1
Putting a = 1 in eq. (i), 1 - b = -1
$\Rightarrow$ -b = -2 $\Rightarrow$ b = 2
Putting a = 1 in eq. (iii), 2 + c = 5
$\Rightarrow$ c = 5 - 2 $\Rightarrow$ c = 3
Putting c = 3 in eq. (iv), 9 + d = 13
$\Rightarrow$ d = 13 - 9 $\Rightarrow$ d = 4
$\Rightarrow$ a = 1, b = 2, c = 3, d = 4
View full question & answer→Question 402 Marks
Find the values of $x, y,$ and $z$ from the following equation:
$\left[\begin{array}{c} {x+y+z} \\ {x+z} \\ {y+z} \end{array}\right]=\left[\begin{array}{l} {9} \\ {5} \\ {7} \end{array}\right]$
Answer$\left[\begin{array}{c} {x+y+z} \\ {x+z} \\ {y+z} \end{array}\right]=\left[\begin{array}{l} {9} \\ {5} \\ {7} \end{array}\right]$
Since, the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding element, we have:
$x + y + z = 9 .... (1)$
$x + z = 5 .... (2)$
$y + z = 7 .... (3)$
Putting the value of equation $(2)$ in equation $(1),$ gives
$y + 5 = 9$
$\Rightarrow y = 4$
Next, putting the value of $y$ in equation $3,$ we get
$4 + z = 7$
$\Rightarrow z = 3$
Finally $, x + z = 5,$ gives
$x + 3 = 5$
$\Rightarrow x = 2$
Therefore $, x = 2, y = 4$ and $z = 3.$
View full question & answer→Question 412 Marks
Find the values of $x,\ y,$ and $z$ from the following equation:
$\left[\begin{array}{cc} {x+y} & {2} \\ {5+z} & {x y} \end{array}\right]=\left[\begin{array}{cc} {6} & {2} \\ {5} & {8} \end{array}\right]$
AnswerSince, the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding element, we have:
$\left[\begin{array}{cc} {x+y} & {2} \\ {5+z} & {x y} \end{array}\right]=\left[\begin{array}{ll} {6} & {2} \\ {5} & {8} \end{array}\right]$
$x + y = 6, xy = 8, 5 + z = 5$
Now, $5 + z = 5$
$\Rightarrow z = 0$
$x+ y = 6 \Rightarrow x = 6 - y$
Also, $xy = 8 \Rightarrow (6 - y)y = 8 \Rightarrow 6y - y^{2}= 8$
$\Rightarrow y^2 - 6y + 8 = 0$
$\Rightarrow y^2 - 4y - 2y + 8 = 0$
$\Rightarrow y(y - 4) - 2(y - 4) = 0$
$\Rightarrow (y - 2)(y - 4) = 0$
Hence, $y = 2$ or $y = 4$
if $y = 2, $ then $x = 6 - 2 = 4$
if $y = 4$, then $x = 6 - 4 = 2$
View full question & answer→Question 422 Marks
Construct a 3 $\times$ 4 matrix, whose elements are given by $a_{ij} = 2i - j$
AnswerIn general $3 \times 4$ matrix is given by $A = \left[\begin{array}{llll} {a_{11}} & {a_{12}} & {a_{13}} & {a_{14}} \\ {a_{21}} & {a_{22}} & {a_{23}} & {a_{24}} \\ {a_{31}} & {a_{32}} & {a_{33}} & {a_{34}} \end{array}\right]$
$a_{ij} = 2i - j, i = 1, 2, 3$ and $j = 1, 2, 3, 4$
Therefore,
$ a_{11}=2 \times 1-1=2-1=1 $
$ a_{21}=2 \times 2-1=4-1=3 $
$ a_{31}=2 \times 3-1=6-1=5 $
$ a_{12}=2 \times 1-2=2-2=0 $
$ a_{22}=2 \times 2-2=4-2=2 $
$ a_{32}=2 \times 3-2=6-2=4 $
$ a_{13}=2 \times 1-3=2-3=-1 $
$ a_{23}=2 \times 2-3=4-3=1 $
$ a_{33}=2 \times 3-3=6-3=3 $
$ a_{14}=2 \times 1-4=2-4=-2 $
$ a_{24}=2 \times 2-4=4-4=0 $
$ a_{34}=2 \times 3-4=6-4=2$
Therefore, required matrix is $A = \left[\begin{array}{cccc} {1} & {0} & {-1} & {-2} \\ {3} & {2} & {1} & {0} \\ {5} & {4} & {3} & {2} \end{array}\right]$
View full question & answer→Question 432 Marks
Construct a 3$\times$4 matrix, whose elements are given by aij $ = \frac{1}{2}\left| { - 3i + j} \right|$
Answer3$\times$4 matrix is given by $A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}} \\ {{a_{21}}} \\ {{a_{31}}} \end{array}\begin{array}{*{20}{c}} {{a_{12}}} \\ {{a_{22}}} \\ {{a_{32}}} \end{array}\begin{array}{*{20}{c}} {{a_{13}}} \\ {{a_{23}}} \\ {{a_{33}}} \end{array}\begin{array}{*{20}{c}} {{a_{14}}} \\ {{a_{24}}} \\ {{a_{34}}} \end{array}} \right]$
Here, ${a_{ij}} = \frac{1}{2}\left| { - 3i + j} \right|$
$\therefore \;{a_{11}} = \frac{1}{2}| - 3 \times 1 + 1| = \frac{1}{2}| - 3 + 1|$ $= \frac{1}{2}| - 2| = \frac{2}{2} = 1$
${a_{21}} = \frac{1}{2}| - 3 \times 2 + 1| = \frac{1}{2}| - 6 + 1| = \frac{1}{2}| - 5| = \frac{5}{2}$
${a_{31}} = \frac{1}{2}| - 3 \times 3 + 1| = \frac{1}{2}| - 9 + 1|$ $ = \frac{1}{2}| - 8| = \frac{8}{2} = 4$
${a_{12}} = \frac{1}{2}| - 3 \times 1 + 2| = \frac{1}{2}| - 3 + 2|$ $ = \frac{1}{2}| - 1| = \frac{1}{2}$
${a_{22}} = \frac{1}{2}| - 3 \times 2 + 2| = \frac{1}{2}| - 6 + 2|$ $ = \frac{1}{2}| - 4| = \frac{4}{2} = 2$
${a_{32}} = \frac{1}{2}| - 3 \times 3 + 2| = \frac{1}{2}| - 9 + 2|$ $= \frac{1}{2}| - 7| = \frac{7}{2}$
${a_{13}} = \frac{1}{2}| - 3 \times 1 + 3| = \frac{1}{2}| - 3 + 3| = 0$
${a_{23}} = \frac{1}{2}| - 3 \times 2 + 3| = \frac{1}{2}| - 6 + 3|$ $= \frac{1}{2}| - 3| = \frac{3}{2}$
${a_{33}} = \frac{1}{2}| - 3 \times 3 + 3| = \frac{1}{2}| - 9 + 3|$ $= \frac{1}{2}| - 6| = \frac{6}{2} = 3$
${a_{14}} = \frac{1}{2}| - 3 \times 1 + 4| = \frac{1}{2}| - 3 + 4| = \frac{1}{2}|1| = \frac{1}{2}$
${a_{24}} = \frac{1}{2}| - 3 \times 2 + 4| = \frac{1}{2}| - 6 + 4|$ $= \frac{1}{2}| - 2| = \frac{2}{2} = 1$
${a_{34}} = \frac{1}{2}| - 3 \times 3 + 4| = \frac{1}{2}| - 9 + 4|$ $= \frac{1}{2}| - 5| = \frac{5}{2}$
Therefore, the required matrix is $A = \left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{5}{2}} \\ 4 \end{array}\;\;\begin{array}{*{20}{c}} {\frac{1}{2}} \\ 2 \\ {\frac{7}{2}} \end{array}\;\;\begin{array}{*{20}{c}} 0 \\ {\frac{3}{2}} \\ 3 \end{array}\;\;\begin{array}{*{20}{c}} {\frac{1}{2}} \\ 1 \\ {\frac{5}{2}} \end{array}} \right]$
View full question & answer→Question 442 Marks
In the matrix $A = \left[\begin{array}{cccc} {2} & {5} & {19} & {-7} \\ {35} & {-2} & {\frac{5}{2}} & {12} \\ {\sqrt{3}} & {1} & {-5} & {17} \end{array}\right]$ write:
- The order of the matrix,
- The number of elements,
- Write the elements $a_{13}, a_{21}, a_{33}, a_{24}, a_{23}.$
Answer
- In the given matrix, the number of rows is $3$ and the number of columns is $4.$
Order of a matrix $=$ No. of rows $\times$ No of columns
Therefore, Order of the given matrix is $3 \times 4.$
- Since, the order of the matrix is $3 \times 4,$ there are $3 \times 4 = 12$ elements in it.
- $a_{13}=19, a_{21}=35, a_{33}=-5, a_{24}=12, a_{23}=\frac{5}{2}$
View full question & answer→Question 452 Marks
Find $x$ and $y$ if $x+y=\left[\begin{array}{ll}5 & 2 \\ 0 & 9\end{array}\right]$ and $x-y=\left[\begin{array}{cc}3 & 6 \\ 0 & -1\end{array}\right]$
Answerx + y + x - y $ = \left[ {\begin{array}{*{20}{c}} 5&2 \\ 0&9 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 3&6 \\ 0&{ - 1} \end{array}} \right]$
$2x = \left[ {\begin{array}{*{20}{c}} 8&8 \\ 0&8 \end{array}} \right]$
$x = \left[ {\begin{array}{*{20}{c}} 4&4 \\ 0&4 \end{array}} \right]$
And, (x + y) - (x - y) $ = \left[ {\begin{array}{*{20}{c}} 5&2 \\ 0&9 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 3&6 \\ 0&{ - 1} \end{array}} \right]$
x + y - x + y $ = \left[ {\begin{array}{*{20}{c}} 2&{ - 4} \\ 0&{10} \end{array}} \right]$
$y = \left[ {\begin{array}{*{20}{c}} 1&{ - 2} \\ 0&5 \end{array}} \right]$
View full question & answer→Question 462 Marks
Find the value of a, b, c and d from the following equation:
$\left[\begin{array}{ll}{2 a+b} & {a-2 b} \\ {5 c-d} & {4 c+3 d}\end{array}\right]$ = $\left[\begin{array}{cc}{4} & {-3} \\ {11} & {24}\end{array}\right]$
Answer$\left[\begin{array}{ll}{2 a+b} & {a-2 b} \\ {5 c-d} & {4 c+3 d}\end{array}\right]$ = $\left[\begin{array}{cc}{4} & {-3} \\ {11} & {24}\end{array}\right]$
As the given matrices are equal, therefore their corresponding elements must be equal.
Comparing the corresponding elements, we get
2a + b = 4 ...(i)
a - 2b = -3 .....(ii)
5c - d = 11 ....(iii)
4c + 3d = 24 ....(iv)
Multiplying (i) by 2 and adding to (ii), we get
5a = 5 $\Rightarrow$ a = 1
$\Rightarrow$ b = 4 - 2(1) = 2
Multiplying (iii) by 3 and adding to (iv), we get
19c = 57 $\Rightarrow$ c = 3
$\Rightarrow$ d = 5(3) - 11 = 4
Hence, a = 1, b = 2, c = 3, d = 4
View full question & answer→Question 472 Marks
If $\left[\begin{array}{ccc}{x+3} & {z+4} & {2 y-7} \\ {-6} & {a-1} & {0} \\ {b-3} & {-21} & {0}\end{array}\right]$ = $\left[\begin{array}{ccc}{0} & {6} & {3 y-2} \\ {-6} & {-3} & {2 c+2} \\ {2 b+4} & {-21} & {0}\end{array}\right]$, Find the values of a, b, c, x, y and z.
AnswerAs the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get
x + 3 = 0,
z + 4 = 6,
2y – 7 = 3y –2
a – 1 = –3,
0 = 2c + 2
b – 3 = 2b + 4
Simplifying, we get a = – 2, b = –7, c = –1, x = –3, y = –5, z = 2
View full question & answer→Question 482 Marks
Construct a 3 $\times$ 2 matrix whose elements are given by $a_{i j}=\frac{1}{2}|i-3 j|$
AnswerIn general a 3 $\times$ 2 matrix is given by $A=\left[\begin{array}{ll} {a_{11}} & {a_{12}} \\ {a_{21}} & {a_{22}} \\ {a_{31}} & {a_{32}} \end{array}\right]$
Now $a_{i j}=\frac{1}{2}$ |i - 3 j|, i = 1, 2, 3 and j = 1, 2.
$\therefore$ $a_{11}=\frac{1}{2}|1-3 \times 1|=1$, $a_{12}=\frac{1}{2}|1-3 \times 2|=\frac{5}{2}$
$a_{21}=\frac{1}{2}|2-3 \times 1|=\frac{1}{2}$, $a_{22}=\frac{1}{2}|2-3 \times 2|=2$
$a_{31}=\frac{1}{2}|3-3 \times 1|=0$, $a_{32}=\frac{1}{2}|3-3 \times 2|=\frac{3}{2}$
Hence the required matrix is given by A = $\left[\begin{array}{cc} {1} & \frac{5}{2} \\ \frac{1}{2} & {2}\\ {0} & {\frac{3}{2}} \end{array}\right]$
View full question & answer→Question 492 Marks
Find AB, if A = $\left[\begin{array}{cc} {0} & {-1} \\ {0} & {2} \end{array}\right] \text { and } B=\left[\begin{array}{ll} {3} & {5} \\ {0} & {0} \end{array}\right]$
AnswerWe have $A B=\left[\begin{array}{rr} {0} & {-1} \\ {0} & {2} \end{array}\right]\left[\begin{array}{ll} {3} & {5} \\ {0} & {0} \end{array}\right]=\left[\begin{array}{ll} {0} & {0} \\ {0} & {0} \end{array}\right]$
Thus, if the product of two matrices is a zero matrix, it is not necessary that one of the matrices is a zero matrix.
View full question & answer→Question 502 Marks
If A = $\left[\begin{array}{rrr} {1} & {-2} & {3} \\ {-4} & {2} & {5} \end{array}\right] \text { and } B=\left[\begin{array}{ll} {2} & {3} \\ {4} & {5} \\ {2} & {1} \end{array}\right]$ , then find AB, BA. Show that AB $\neq$ BA
AnswerSince A is a 2 $\times$ 3 matrix and B is 3 $\times$ 2 matrix.
Hence AB and BA are both defined and are matrices of order 2 $\times$ 2 and 3 $\times$ 3, respectively.
Now,$A B=\left[\begin{array}{rrr} {1} & {-2} & {3} \\ {-4} & {2} & {5} \end{array}\right]\left[\begin{array}{ll} {2} & {3} \\ {4} & {5} \\ {2} & {1} \end{array}\right]$ = $\left[\begin{array}{cc} {2-8+6} & {3-10+3} \\ {-8+8+10} & {-12+10+5} \end{array}\right]$ = $\left[\begin{array}{rr} {0} & {-4} \\ {10} & {3} \end{array}\right]$
and $B A=\left[\begin{array}{cc} {2} & {3} \\ {4} & {5} \\ {2} & {1} \end{array}\right]\left[\begin{array}{ccc} {1} & {-2} & {3} \\ {-4} & {2} & {5} \end{array}\right]$ = $\left[\begin{array}{ccc} {2-12} & {-4+6} & {6+15} \\ {4-20} & {-8+10} & {12+25} \\ {2-4} & {-4+2} & {6+5} \end{array}\right]$ = $\left[\begin{array}{ccc} {-10} & {2} & {21} \\ {-16} & {2} & {37} \\ {-2} & {-2} & {11} \end{array}\right]$
Clearly AB $\neq$ BA
In the above example, both AB and BA are of different orders and so AB $\neq$ BA.
View full question & answer→Question 512 Marks
Find AB, if $A=\left[\begin{array}{ll} {6} & {9} \\ {2} & {3} \end{array}\right] \text { and } B=\left[\begin{array}{lll} {2} & {6} & {0} \\ {7} & {9} & {8} \end{array}\right]$
AnswerMatrix A has 2 columns which is equal to the number of rows of B. Hence AB is defined. Now
$A B=\left[\begin{array}{lll} {6(2)+9(7)} & {6(6)+9(9)} & {6(0)+9(8)} \\ {2(2)+3(7)} & {2(6)+3(9)} & {2(0)+3(8)} \end{array}\right]$
$=\left[\begin{array}{ccc} {12+63} & {36+81} & {0+72} \\ {4+21} & {12+27} & {0+24} \end{array}\right]$
$=\left[\begin{array}{ccc} {75} & {117} & {72} \\ {25} & {39} & {24} \end{array}\right]$
View full question & answer→Question 522 Marks
Consider the following information regarding the number of men and women workers in three factories I, II and III | | Men workers | Women workers |
| I | 30 | 25 |
| II | 25 | 31 |
| III | 27 | 26 |
Represent the above information in the form of a 3 $\times$ 2 matrix. What does the entry in the third row and second column represent?
AnswerThe information is represented in the form of a 3$\times$2 matrix as follows
$A=\left[\begin{array}{ll} {30} & {25} \\ {25} & {31} \\ {27} & {26} \end{array}\right]$
The entry in the third row and the second column represents the number of women workers in factory III.
View full question & answer→