CBSE Test Paper 01
Chapter 3 Matrices
If A=⎡⎣⎢15−212−136−3⎤⎦⎥. Then |A| is
- none of these
- Idempotent
- Nilpotent
- Symmetric
∣∣∣∣1e210212–√2∣∣∣∣is equal to
A square matrix A is called idempotent if
Let for any matrix M ,M-1 exist. Which of the following is not true.
- none of these
- (M-1)-1 = M
- (M-1)2 = (M2)-1
- (M-1)-1 = (M-1)1
The system of equations x + 2y = 11, -2 x – 4y = 22 has
- only one solution
- infinitely many solutions
- finitely many solutions
- no solution
- Sum of two skew-symmetric matrices is always ________ matrix.
- ________ matrix is both symmetric and skew-symmetric matrix.
- If A and B are square matrices of the same order, then (kA)' = ________ where k is any scalar.
If A=[2435], Prove that A – At is a skew – symmetric matrix.
A=[2546], Prove that A + A' is a symmetric matrix.
The no. of all possible matrics of order 3 × 3 with each entry as 0 or 1 is-
Find the matrix X so that X[123456]=[−7−8−9246].
If A is a square matrix, such that A2 = A, then (I + A)3 – 7A is equal to
If A=[2546], then Prove that A + A' is a symmetric matrix.
If f(x) = x2 – 5x + 7 and A=[3−112] then find f(A).
Find the matrix A such that ⎡⎣⎢2−110−34⎤⎦⎥A=⎡⎣⎢−119−8−222−10−515⎤⎦⎥.
Show that:
- [56−17][2314]≠[2314][56−17]
- ⎡⎣⎢101211300⎤⎦⎥⎡⎣⎢−1021−13014⎤⎦⎥ ≠⎡⎣⎢−1021−13014⎤⎦⎥⎡⎣⎢101211300⎤⎦⎥
A=[0tanα2−tanα20],
Prove I+A=(I−A)[cosαsinα−sinαcosα].
CBSE Test Paper 01
Chapter 3 Matrices
Solution
- Nilpotent
Explanation: The given matrix A is nilpotent, because |A| = 0,as determinant of a nilpotent matrix is zero.
- 0
Explanation: ∣∣∣∣1e210212–√2∣∣∣∣=2∣∣∣∣1e110112–√1∣∣∣∣=0, because, row 1 and row 3 are identical.
- A2 = A
Explanation: If the product of any square matrix with itself is the matrix itself, then the matrix is called Idempotent.
- (M-1)-1 = (M-1)1
Explanation: Clearly , (M-1)-1 = (M-1)1 is not true.
- no solution
Explanation: For no solution , we have: a1a2=b1b2≠c1c2, for given system of equations we have: 1−2=2−4≠1122.
- skew symmetric
Null
- kA'
- P = A - At
=[2435]+[−2−3−4−5]
=[01−10]
P′=[0−110]
P′=−[01−10]
P' = - P
Proved. - P=A+A′=[2546]+[2456]
P=[49912]
P′=[49912]
Thus, P' = P. - 29=512
- Let X=[acbd]
∴[acbd][123456]=[−7−8−9246]
[a+4b2a+5b3a+6bc+4d2c+5d3c+6d] =[−7−8−9246]
Hence, a = 1, b = -2, c = 2, d = 0
X=[12−20] - (I + A)3 - 7A = I3 + A3 + 3IA (I + A) - 7A
= I + A3 + 3I2A + 3IA2 - 7A
= I + A3 + 3A + 3A2 - 7A
= I + A3 + 3A + 3A - 7A {A2 = A}
= I + A3 - A
= I + A2 - A [A2 = A, A3 = A2]
= I + A - A {A2 = A}
= I - P = A + A' =[2546]+[2456]
P=[49912]
P′=[49912]
P' = P
Therefore P = P'
Hence A+A' is symmetric.
A=[3−112]
A2=AA=[3−112][3−112]=[8−553]
Now, f(A) = A2 - 5A + 7I
=[8−553]−5[3−112]+7[1001]
f(A)=[8−15+7−5+5+05−5+03−10+7]=[0000]
Therefore, f(A)=[0000]
- We have, ⎡⎣⎢2−110−34⎤⎦⎥A=⎡⎣⎢−119−8−222−10−515⎤⎦⎥
From the given equation, it is clear that order of A should be 2×3
Let A=[abcdef]
∴ ⎡⎣⎢21−3−104⎤⎦⎥[adbecf]=⎡⎣⎢−119−8−222−10−515⎤⎦⎥
⇒⎡⎣⎢2a−da+0d−3a+4d2b−eb+0.e−3b+4e2c−fc+0.f−3c+4f⎤⎦⎥ =⎡⎣⎢−119−8−222−10−515⎤⎦⎥
⇒⎡⎣⎢2a−da−3a+4d2b−eb−3b+4e2c−fc−3c+4f⎤⎦⎥ =⎡⎣⎢−119−8−222−10−515⎤⎦⎥
By equality of matrices, we get
a = 1, b = -2, c = -5
and 2a - d = -1 ⇒ d = 2a + 1 = 3;
⇒ 2b - e = -8 ⇒ e = 2(-2) + 8 =4
2c - f = -10 ⇒ f = 2c + 10 = 0
∴A=[1−2−5340] - L.H.S. =[56−17][2314] =[5(2)+(−1)36(2)+7(3)5(1)+(−1)46(1)+7(4)]=[733134]
R.H.S. =[2314][56−17] =[2(5)+1(6)3(5)+4(6)2(−1)+1(7)3(−1)+4(7)]=[1639525]
∴ L.H.S. ≠ R.H.S. - L.H.S. =⎡⎣⎢101211300⎤⎦⎥⎡⎣⎢−1021−13014⎤⎦⎥
=⎡⎣⎢1(−1)+2(0)+3(2)0(−1)+1(0)+0(2)1(−1)+1(0)+0(2)1(1)+2(−1)+3(3)0(1)+1(−1)+0(3)1(1)+1(−1)+0(3)1(0)+2(1)+3(4)0(0)+1(1)+0(4)1(0)+1(1)+0(4)⎤⎦⎥
=⎡⎣⎢50−18−101411⎤⎦⎥
R.H.S. =⎡⎣⎢−1021−13014⎤⎦⎥⎡⎣⎢101211300⎤⎦⎥
=⎡⎣⎢−1(1)+1(0)+0(1)0(1)+(−1)0+1(1)2(1)+3(0)+4(1)(−1)2+1(1)+0(1)(0)2+1(−1)+1(1)2(2)+3(1)+4(1)(−1)3+1(0)+0(0)(0)3+0(−1)+1(0)2(3)+3(0)+4(0)⎤⎦⎥
=⎡⎣⎢−116−1011−306⎤⎦⎥
∴ L.H.S. ≠ R.H.S.
- Put tanα2=t
A=[0t−t0]
I+A=[1001]+[0t−t0]
=[1t−t1]
I−A=[1001]−[0t−t0]
=[1001]+[0−tt0]
=[1−tt1]
L.H.S. =(I−A)[cosαsinα−sinαcosα]
=(I−A)⎡⎣⎢⎢1−tan2α21+tan2α22tan2α21+tan2α2−2tan2α21+tan2α21−tan2α21+tan2α2⎤⎦⎥⎥
=[1−tt1]⎡⎣⎢1−t21+t2−2t1+t2−2t1+t21−t21+t2⎤⎦⎥
=⎡⎣⎢1−t21+t2+t.2t1+t2−t(1−t21+t2)+2t1+t2−2t1+t2+t(1−t21+t2)−t(−2t1+t2)+(1−t21+t2)⎤⎦⎥
=⎡⎣⎢1−t2+2t21+t2−t+t3+2t1+t2−2t+t−t31+t22t2+1−t21+t2⎤⎦⎥
=⎡⎣⎢1+t21+t2t3+t1+t2−t3−t1+t2t2+11+t2⎤⎦⎥
=⎡⎣⎢1t(1+t2)1+t2−t(1+t2)1+t2t2+11+t2⎤⎦⎥
=[1t−t1]
L.H.S = R.H.S
Hence proved